NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations - Exercise 9.5
Access Exercises of Class 12 Maths Chapter 9 – Differential Equations
Exercise 9.1 Solutions 12 Questions
Exercise 9.2 Solutions 12 Questions
Exercise 9.3 Solutions 12 Questions
Exercise 9.4 Solutions 23 Questions
Exercise 9.5 Solutions 17 Questions
Exercise 9.6 Solutions 19 Questions
Miscellaneous Exercise on Chapter 9 Solutions 18 Questions
Exercise 9.5
Direction (Q. 1 to 10) : In each of the question show that the given differential equation is homogeneous and solve each of them.
1. (x2 + xy) dy = (x2 + y2) dx.
Sol. Given, differential equation is
$$\frac{dy}{dx} = \frac{\text{x}^{2} + y^{2}}{x^{2} + xy}\space\text{...(i)}$$
Here, numerator and denominator both have polynomial of degree 2. So, the given differential equation is homogeneous.
Let y = vx
$$\Rarr\space\frac{dy}{dx} = v + x\frac{dv}{dx}\\\text{On putting the values of}\space\frac{dy}{dx}\\\text{and y in eq. (i), we get}\\ v + x\frac{dv}{dx} = \frac{x^{2} + x^{2}v^{2}}{x^{2} + vx^{2}}\\\Rarr\space v+x\frac{dv}{dx} = \frac{x^{2}(1+v^{2})}{x^{2}(1+v)}\\\Rarr\space x\frac{dv}{dx} = \frac{1 + v^{2}}{1+v}-v\\\Rarr\space x\frac{dv}{dx} = \frac{1 + v^{2} - v - v^{2}}{1+v}\\=\frac{1-v}{1+v}$$
$$\Rarr\space \frac{1+v}{1-v}dv = \frac{dx}{x}\\\Rarr\space \frac{2-1+v}{1-v}dv = \frac{dx}{x}\\\Rarr\space \bigg(\frac{2}{1-v} -\frac{1-v}{1-v}\bigg)dv = \frac{dx}{x}\\\Rarr\bigg(\frac{2}{1-v}-1\bigg)dv=\frac{dx}{x}$$
On integrating both sides, we get
$$\int\frac{2}{1-v}dv - \int 1 \space dv = \int\frac{1}{x}dx\\\Rarr\space -2\space\text{log}(1-v)-v = \text{log x - log C}\\\Rarr\space -v-2\space\text{log}(1-v) - \text{log x} = -\text{log}\space C\\\Rarr\space v+2\space\text{log}(1-v) + \text{log x = log C}\\\text{Put v =}\frac{y}{x}\\\Rarr\space \frac{y}{x} + 2\space\text{log}\bigg(1-\frac{y}{x}\bigg)+\\\text{log x = log C}\\\Rarr\space\frac{y}{x} + \text{log}\bigg(\frac{x-y}{x}\bigg)^{2}+\\\text{log x = log C}$$
$$\Rarr\space\frac{y}{x} + \text{log}(x-y)^{2}-\text{log\space x}^{2} +\\ \text{log x = log C}\\\Rarr\space\frac{y}{x} + \text{log}(x-y)^{2} - \\2\space\text{log x + log x = log C}\\\Rarr\space\frac{y}{x} + \text{log}(x-y)^{2} - \text{log x = log C}\\\Rarr\space \frac{y}{x}+\text{log}\frac{(x-y)^{2}}{x}-\text{log C = 0}\\\Rarr\space \text{log}\bigg(\frac{(x-y)^{2}}{x}.\frac{1}{\text{C}}\bigg) = \frac{y}{x}\\\Rarr\space (x-y)^{2} = \text{Cxe}^{-\frac{y}{x}}$$
This is the required solution of the given differential equation.
$$\textbf{2.\space y'} =\frac{\textbf{x+y}}{\textbf{x}}\textbf{.}\\\textbf{Sol.\space}\text{Given\space} y' = \frac{dy}{dx} = \frac{x+y}{x}\space\text{...(i)}$$
Here, the given differential equation is homogeneous.
$$\text{So, put\space} y = vx\\\Rarr\space\frac{dy}{dx} = v+x\frac{dv}{dx}$$
$$\text{On putting values of}\frac{dy}{dx}\\\text{and y in eq. (i), we get}\\\therefore\space\text{v+x}\frac{dv}{dx} = \frac{x + vx}{x}\\\Rarr\space v + x\frac{dv}{dx} = 1+v\\\Rarr\space \frac{dv}{dx}=\frac{1}{x}\\\Rarr\space dv = \frac{dx}{x}$$
On integrating both sides, we get
$$\int dv = \int\frac{1}{x}dx\\\Rarr\space v = \text{log}|x|+\text{C}\\\Rarr\space \frac{y}{x} = \text{log}|x|+\text{C}\\\lbrack\because v =\frac{y}{x}\rbrack\\\Rarr\space \text{y = x log}|x| \text{+ Cx.}$$
This is the required solution of the given differential equation.
3. (x – y) dy – (x + y) dx = 0.
Sol. Given, (x – y)dy – (x + y)dx = 0
$$\Rarr\space (x-y)dy = (x+y)dx\\\Rarr\space \frac{dy}{dx} = \frac{\text{x+y}}{\text{x-y}}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
So put y = vx
$$\Rarr\space\frac{dy}{dx} = v + \frac{dv}{dx}$$
On putting values of
$$\frac{dy}{dx}\space\text{and y in eq. (i), we get}$$
$$\text{v+x}\frac{dv}{dx} = \frac{x+vx}{x-vx}\\\Rarr\space v+x\frac{dv}{dx} = \frac{\text{1+v}}{\text{1-v}}\\\Rarr\space x\frac{dv}{dx} = \frac{\text{1+v}}{\text{1-v}}-v\\\Rarr\space x\frac{dv}{dx}=\frac{1+v-v+v^{2}}{\text{1-v}}\\\Rarr\space \frac{\text{1-v}}{\text{1+v}^{2}}dv = \frac{dx}{x}\\\int\frac{\text{1-v}}{\text{1+v}^{2}}\text{dv} = \int\frac{dx}{x}$$
On integrating both sides, we get
$$\Rarr\space \int\frac{1}{1 + v^{2}}dv - \int\frac{v}{\text{1+v}^{2}}dv\\=\text{log}|x|+\text{C}\\\text{Let\space 1 + v}^{2} = t\\\Rarr 2v = \frac{dt}{dv}\\\Rarr\space dv = \frac{dt}{2v},\\\therefore\space\text{tan}^{\normalsize-1}v - \int\frac{v}{t}×\frac{dt}{2v} \\= \text{log}|x| + \text{C}\\\Rarr\space \text{tan}^{\normalsize-1}v-\frac{1}{2}\text{log}|t|\\=\text{log}|x| + \text{C}\\\Rarr\space 2 \space\text{tan}^{\normalsize-1}v - \lbrack\text{log}(1+v^{2}) + 2 \text{log}(x)\rbrack = 2\text{C}$$
(Put t = 1 + v2)
$$\Rarr\space 2\space\text{tan}^{\normalsize-1}v - \text{log}\lbrack(1+v^{2}x^{2})\rbrack = 2\text{C}\\\Rarr\space \text{2 tan}^{\normalsize-1}x\frac{y}{x}- \text{log}\bigg[\bigg(\frac{x^{2}+y^{2}}{x^{2}}\bigg)x^{2}\bigg]= 2\text{C}$$
$$\bigg(\text{Put v}= \frac{y}{x}\bigg)\\\Rarr\space \text{2 tan}^{\normalsize-1}\frac{y}{x} - \text{log}(x^{2}+y^{2}) = 2\text{C}\\\Rarr\space \text{tan}^{\normalsize-1}\frac{y}{x} - \frac{1}{2}\text{log}(x^{2} + y^{2}) = \text{C}$$
This is the required solution of the given differential equation.
$$\textbf{4.\space}\textbf{(x}^{\textbf{2}} \textbf{- y}^{\textbf{2}}\textbf{)dx + 2xy dy = 0.}$$
Sol. Given, (x2 – y2) dx + 2xydy = 0
$$\Rarr\space 2xy\space dy = (y^{2}-x^{2})dx\\\Rarr\space \frac{dy}{dx}= \frac{y^{2}-x^{2}}{2xy}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
So, put y = vx
$$\Rarr\space \frac{dy}{dx} = v + x\frac{dv}{dx},\\\text{then Eq. (i) becomes}$$
$$v + x\frac{dv}{dx} = \frac{v^{2}x^{2}-x^{2}}{2x^{2}v}\\\Rarr\space v + x\frac{dv}{dx} = \frac{v^{2}-1}{2v}\\\Rarr\space x \frac{dv}{dx} = \frac{v^{2}-1}{2v}-v\\\Rarr\space x\frac{dv}{dx} = \frac{-1-v^{2}}{2v}\\\Rarr\space \frac{2v}{v^{2}+1}dv = -\frac{dx}{x} $$
On integrating both sides, we get
$$\int\frac{2v}{v^{2}+1}dv + \int\frac{dx}{x} = 0\\\text{Let\space v}^{2}+1= t\\\Rarr 2v = \frac{dt}{dv}\Rarr\space dv = \frac{dt}{2v}.\\\therefore\space \int\frac{2v}{t}×\frac{dt}{2v}+\int\frac{dx}{x} = 0$$
$$\Rarr\space \text{log}|t| + \text{log}|x| = \text{log C}\\\Rarr\space \text{log}|(v^{2}+1)x| = \text{log C}\\(\because\space t = 1 + v^{2})\\\Rarr\space \text{log}\bigg[\bigg(\frac{y^{2}+x^{2}}{x^{2}}\bigg)x\bigg] = \text{log C}\\\Rarr\space \frac{y^{2}+x^{2}}{x}+\text{C}\\(\because\space v=\frac{y}{x}) \\\Rarr\space x^{2}+ y^{2} = \text{Cx}.$$
This is the required solution of the given differential equation.
$$\textbf{5.\space x}^{\textbf{2}}\frac{\textbf{dy}}{\textbf{dx}}\textbf{ = x}^{\textbf{2}}\textbf{-2y}^{\textbf{2}}\textbf{ + xy.}\\\textbf{Sol.\space}\text{Given,\space}\frac{dy}{dx} = \frac{x^{2}- 2y^{2} + xy}{x^{2}}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
So, put y = vx
$$\Rarr\space \frac{dy}{dx} = v + x\frac{dv}{dx}\\\text{Then, eq. (i) becomes}\\\text{v+x}\frac{dv}{dx} = \frac{x^{2}-2x^{2}v^{2} + x^{2}v}{x^{2}}\\\Rarr\space v+x\frac{dv}{dx} = 1-2v^{2}+v\\\Rarr\space x\frac{dv}{dx} = 1 - 2v^{2}\\\Rarr\space \frac{1}{1-2v^{2}}dv = \frac{1}{x}dx$$
On integrating both sides, we get
$$\int\frac{1}{1 - 2v^{2}}dv = \int\frac{dx}{x}\\\Rarr\space \frac{1}{2}\int\frac{1}{\bigg(\frac{1}{\sqrt{2}}\bigg)^{2}-v^{2}}dv = \int\frac{dx}{x}\\\Rarr\space \frac{1}{2}.\frac{1}{2.\frac{1}{\sqrt{2}}}\text{log}\begin{vmatrix}\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\end{vmatrix} = \text{log}|x| + \text{C}\\\bigg[\because\space \int\frac{dx}{a^{2}-x^{2}} = \frac{1}{2a}\text{log}\bigg|\frac{a+x}{a-x}\bigg| + \text{C}\bigg]\\\Rarr\space \frac{1}{2\sqrt{2}}\text{log}\bigg|\frac{1 + \sqrt{2}v}{1 - \sqrt{2}v}\bigg| = \text{log}|x| + \text{C}\\\Rarr\space \frac{1}{2\sqrt{2}}\text{log}\bigg|\frac{1 + \sqrt{2}\frac{y}{x}}{1 - \sqrt{2}\frac{y}{x}}\bigg|\\=\text{log}|x| + \text{C}$$
This is the required solution of given differential equation.
$$\textbf{6.\space xdy - y dx = }\sqrt{\textbf{x}^{\textbf{2}} \textbf{+ y}^{\textbf{2}}}\space\textbf{dx.}\\\textbf{Sol.\space}\text{Given, \space xdy}\\ = y dx +\sqrt{x^{2} + y^{2}}dx\\\Rarr\space \frac{dy}{dx} = \frac{y +\sqrt{x^{2} + y^{2}}}{x}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
So, put y = vx
$$\Rarr\space \frac{dy}{dx} = v+x\frac{dv}{dx},\\\text{then eq. (i) becomes}\\\text{v+x}\frac{dv}{dx} = \frac{vx + \sqrt{x^{2} + x^{2}v^{2}}}{x}\\\Rarr\space v+x\frac{dv}{dx}=\\\frac{vx + \sqrt{x^{2} + x^{2}v}}{x}\\\Rarr\space \text{v + x}\frac{dv}{dx} = \frac{x\bigg[ v + \sqrt{1 + v^{2}}\bigg]}{x}\\\Rarr\space \text{v + x}\frac{dv}{dx} = v + \sqrt{1 + v^{2}}\\\Rarr\space \frac{1}{\sqrt{1+v^{2}}}dv = \frac{1}{x}dx$$
On integrating both sides, we get
$$\int\frac{1}{\sqrt{1+v^{2}}}dv = \int\frac{dx}{x}\\\text{log}(v + \sqrt{1 + v^{2}}) = \text{log x + C}\\\bigg[\because\space \int\frac{dx}{\sqrt{x^{2} + a^{2}}} = \text{log}|x + \sqrt{x^{2} + a^{2}}|+\text{C}\bigg]\\\Rarr\space\text{log}\bigg[\frac{y}{x} + \sqrt{1+ \bigg(\frac{y}{x}\bigg)^{2}}\bigg]\\=\text{log x + log C}\\\bigg(\text{Put v} = \frac{y}{x}\bigg)\\\Rarr\space\text{log}\bigg[\frac{y}{x} + \sqrt{\frac{x^{2} + y^{2}}{x^{2}}}\bigg] = \text{log (x C)}$$
$$\lbrack\because\space \text{log m + log n = log mn}\rbrack\\\Rarr\space \frac{y}{x} + \frac{\sqrt{x^{2}+y^{2}}}{x}= \text{Cx}\\\lbrack\because\space \text{log m = log m}\Rarr\space mn\rbrack\\\Rarr\space y + \sqrt{x^{2} + y^{2}} = \text{Cx} ^{2}$$
which is the required solution.
$$\textbf{7.\space}\begin{Bmatrix}\textbf{x cos}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg) \textbf{+ y sin x}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg)\end{Bmatrix}\textbf{y\space dx}\\\textbf{=}\begin{Bmatrix}\textbf{y sin}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg) - \textbf{x cos}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg)\end{Bmatrix}\textbf{x dy.}\\\textbf{Sol.\space}\text{Given, }\begin{Bmatrix}\text{x cos}\bigg(\frac{y}{x}\bigg) + \text{y sin}\bigg(\frac{y}{x}\bigg)\end{Bmatrix}y dx\\=\begin{Bmatrix}\text{y sin}\bigg(\frac{y}{x}\bigg) - \text{x cos}\bigg(\frac{y}{x}\bigg)\end{Bmatrix}x\space dy\\\Rarr\space\frac{dy}{dx} = \frac{y\begin{Bmatrix}\text{x cos}\frac{y}{x} + \text{y sin}\frac{y}{x}\end{Bmatrix}}{x\begin{Bmatrix}\text{y sin}\frac{y}{x} - \text{x cos }\frac{y}{x}\end{Bmatrix}}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
So, put y = vx
$$\Rarr\space\frac{dy}{dx} = v + x\frac{dv}{dx}\\\therefore\space \text{v + x}\frac{dv}{dx} = \frac{vx\lbrace\text{x cos v + vx sin v}\rbrace}{x\lbrace vx\space sin v - x\space cos v\rbrace}\\\Rarr\space v +x\frac{dv}{dx} = \frac{v\lbrace cos\space v + sin \space v\rbrace}{v \space sin v - cos v}\\\Rarr\space x\frac{dv}{dx} = \frac{v\space \text{cos v} + v^{2} sin\space v}{v \space sin v - cos\space v}-v\\\Rarr\space x\frac{dv}{dx} = \frac{\text{v cos v + v}^{2} \text{sin \space v} - v^{2}\space\text{sin v + v cos v}}{\text{v \space sin v - cos v}}\\\Rarr\space x\frac{dv}{dx} = \frac{\text{2v cos v}}{\text{v\space sin v - cos v}}\\\Rarr\space \bigg(\frac{\text{v sin v - cos v}}{\text{v cos v}}\bigg)dv = \frac{2}{x}dx$$
$$\Rarr\space\bigg(\text{tan v} - \frac{1}{v}\bigg)dv = \frac{2}{x}dx$$
On integrating both sides, we get
$$\int\bigg(\text{tan v - }\frac{1}{v}\bigg)dv = \int\frac{2 \space dx}{x}\\\Rarr\space \int\text{tan v dv} - \int\frac{1}{v}dv = 2\int\frac{1}{x}dx$$
$$\Rarr\space -\text{log}|\text{cos v}|- \text{log}|v| = 2\space\text{log}|x|+ \text{C}\\\Rarr\space\text{log}|\text{v\space cos v}| + 2\space\text{log}|x|=-\text{C}\\(\because\space \text{log m + log n = log mn})\\\Rarr\space \text{log}\lbrack(\text{v cos v})x^{2}\rbrack = -\text{C}\\\Rarr\space (v \space cos v)x^{2} = e^{-c}\\\lbrack\because\space \text{log}_e x = m \Rarr\space e^{m}=x\rbrack\\\Rarr\space x^{2}v\text{cos v = A}\\(\text{where, A = e}^{-C})\\\Rarr\space x^{2}\frac{y}{x}\text{cos}\frac{y}{x} = A\\\bigg(\because\space y = \frac{v}{x}\bigg)\\\Rarr\space xy\space cos\frac{y}{x}= A$$
This is the required solution of the given differential equation.
$$\textbf{8.\space x}\frac{\textbf{dy}}{\textbf{dx}}\textbf{-y} \textbf{+ x sin}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg) \textbf{= 0.}\\\textbf{Sol.\space}\text{Given,}\space\frac{dy}{dx} = \frac{y}{x}-\text{sin}\frac{y}{x}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So,\space Put}\space\frac{y}{x} = v\\\Rarr\space y = vx\\\Rarr\space \frac{dy}{dx} = v + x\frac{dv}{dx}$$
Then, eq. (i) becomes
$$v + x\frac{dv}{dx} \text{= v - sin\space v}\\\Rarr\space \text{cosec v dv} = \frac{1}{x}dx$$
On integrating both sides, we get
$$\int\text{cosec v dv =} -\int\frac{dx}{x}\\\Rarr\space \text{log}|\text{cosec v - cot v}|= -\text{log}|x| + A\\(\because\space\int\text{cosec x dx = log |\text{cosec x - cot x}|})\\\Rarr\space\text{log}|\text{cosec v - cot v}| + \text{log}|x| = A\\\Rarr\space |x\space \text{cosec v - cot v}| = e^{A}\\\Rarr\space x\bigg(\frac{1}{\text{sin v}}-\frac{\text{cos v}}{\text{sin v}}\bigg) = e^{\text{A}}\\\Rarr\space x\bigg(\frac{\text{1 - cos v}}{\text{sin v}}\bigg) = e^{\text{A}}\\\Rarr\space x(\text{1 - cos v}) = e^{A}\space\text{sin v}\\\Rarr\space x\bigg(\text{1 - cos}\bigg(\frac{y}{x}\bigg)\bigg) = \text{C sin}\bigg(\frac{y}{x}\bigg)$$
$$\bigg(\text{Put C = e}^{A}\text{and v} = \frac{y}{x}\bigg)$$
This is the required solution of given differential equation.
$$\textbf{9.\space y dx + x log}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg)\textbf{dy}\textbf{- 2x dy = 0.}$$
$$\textbf{Sol.\space}\text{Given,\space y dx + x log}\bigg(\frac{y}{x}\bigg)dy-\\\text{2x dy = 0}\\\Rarr\space\frac{y}{x} + \text{log}\bigg(\frac{y}{x}\bigg)\frac{dy}{dx}-2\frac{dy}{dx}=0\\\Rarr\space \frac{dy}{dx} = \frac{\frac{y}{x}}{\text{2 - log}\frac{y}{x}}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So, put}\frac{y}{x} = v\space\text{i.e., y = vx}\\\Rarr\space\frac{dy}{dx} = v + x\frac{dv}{dx}\\\text{Then, eq. (i) becomes}\\\text{v + x}\frac{dv}{dx} = \frac{v}{2 -\text{log v}}\\\Rarr\space x\frac{dv}{dx} = \frac{v}{\text{2 - log v}}-v\\\Rarr\space x\frac{dv}{dx} = \frac{v - 2v + v log v}{\text{2 - log v}}\\\Rarr\space x \frac{dv}{dx} = \frac{\text{v log v - v}}{\text{2 - log v}}\\\Rarr\space \frac{\text{2 - log v}}{\text{v log v-v}}dv = \frac{1}{x}dx$$
$$\Rarr\space\frac{\text{1 - (log v - 1)}}{\text{v(log v-1)}}dv = \frac{1}{x}dx\\\Rarr\space\bigg(\frac{1}{\text{v(log v -1)}}- \frac{\text{(log v-1)}}{v(\text{log v-1})}\bigg)dv = \frac{1}{x}dx$$
On integrating both sides, we get
$$\int\bigg(\frac{1}{v\text{(log v-1)}}-\frac{1}{v}\bigg)dv = \int\frac{dx}{x}\\\Rarr\space\int\frac{1}{\text{v(log v-1)}}dv - \int\frac{1}{v}dv\\=\int\frac{dx}{x}\\\text{Let log v-1 = t}\\\Rarr\space\frac{1}{v}dv = dt\\\therefore\space\int\frac{dt}{t}-\int\frac{1}{v}dv = \int\frac{dx}{x}$$
$$\Rarr\space \text{log}|t| - \text{log}|v| = \text{log}|x| + \text{log C}\\\Rarr\space \text{log}|\text{log v-1}| - \text{log}|v| =\text{log}|x| + \text{log C}\\\lbrack\because\space \text{t = log v - 1}\rbrack\\\Rarr\space\text{log}|v-1| - \text{log}|v| - \text{log}|x| = \text{log C}\\\Rarr\space \text{log}\bigg|\frac{\text{log v-1}}{vx}\bigg| = \text{log C}\\\bigg[\because\space\text{log}\frac{m}{n} = \text{log m - log n}\bigg]\\\Rarr\space\bigg|\frac{\text{log v - 1}}{\text{vx}}\bigg| = \text{log C}\\\Rarr\space \frac{\text{log v - 1}}{vx} = \text{C}$$
$$\Rarr\space\frac{\text{log}\bigg(\frac{y}{x}\bigg)-1}{y} = \text{C}\\\bigg[\because\space v = \frac{y}{x}\bigg]\\\Rarr\space\text{log}\bigg(\frac{y}{x}\bigg)-1 = \text{Cy}$$
This is the required solution of the given differential equation.
$$\textbf{10.\space}(\textbf{1 + e}^{\textbf{x/y}})\textbf{dx +}\\ \textbf{e}^{\frac{\textbf{x}}{\textbf{y}}}\bigg(\textbf{1 - }\frac{\textbf{x}}{\textbf{y}}\bigg)\textbf{dy = 0.}\\\textbf{Sol.}\space\text{Given}\space (1 + e^{\frac{x}{y}})dx = e^{x/y}\bigg(\frac{x}{y}-1\bigg)dy$$
$$\Rarr\space\frac{dy}{dx} = \frac{e^{\frac{x}{y}}\bigg(\frac{x}{y}-1\bigg)}{e^{\frac{x}{y}}+1}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So, put}\space\frac{x}{y} = v\space\text{i.e., x = vy}\\\Rarr\space\frac{dx}{dy} = v+y\frac{dv}{dy}\\\text{Then, eq. (i) becomes}\\v + y\frac{dv}{dy} = \frac{e^{v}(v-1)}{e^{v}+1}\\\Rarr\space y\frac{dv}{dy} = \frac{e^{v}(v-1)}{e^{v}+1}-v\\\Rarr\space y\frac{dv}{dy} = \frac{ve^v - e^v - ve^v-v}{e^v+1}\\\Rarr\space \frac{e^v+1}{e^v+v}dv = -\frac{1}{y}dy$$
On integrating both sides, we get
$$\int\frac{e^{v}+1}{e^v+v}dv = -\int\frac{1}{y}dy\\\text{Put e}^v + v=t\\\Rarr\space e^v + 1 = \frac{dt}{dv}\\\Rarr\space dv = \frac{dt}{e^v+1}\\\int\frac{e^v+1}{t}×\frac{dt}{e^v+1} \\= -\text{log}|y| + \text{log C}\\\Rarr\space\text{log}|t| + \text{log}|y| = \text{log C}\\\Rarr\space \text{log}|e^v + v| + \text{log}|y| = \text{log C}\\\lbrack\because\space t = e^v + v \rbrack\\\Rarr\space \text{log}|(e^v + v)y| = \text{log C} $$
$$\Rarr\space |(e^v + v)y| \text{= C}\\\Rarr\space (e^v + v)y \text{= C}\\\text{So, put v} = \frac{x}{y},\text{we get}\\\bigg(e^{\frac{x}{y}}+\frac{x}{y}\bigg)y = \text{C}\\\Rarr\space ye^{\frac{x}{y}} + x = \text{C}$$
This is the required solution of the given differential equation.
Direction (Q. 11 to 15) : For each of the differential equation find the particular solution satisfying the given condition.
11. (x + y)dy + (x – y)dx = 0, y = 1 when x = 1.
$$\text{Given, (x + y)dy} = (y -x)dx\\\Rarr\space \frac{dy}{dx} = \frac{\text{y-x}}{\text{x + y}}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So, put}\frac{y}{x} = v\space\text{i.e., y = vx}\\\Rarr\space\frac{dy}{dx} = v+x\frac{dv}{dx}\\\text{Then, eq, (i) becomes}\\\text{v + x}\frac{dv}{dx} = \frac{vx-x}{x + vx}\\\Rarr\space v+x\frac{dv}{dx} = \frac{x(v-1)}{x(1 - v)}\\\Rarr\space v + x\frac{dv}{dx} = \frac{v-1}{1 + v}\\\Rarr\space x\frac{dv}{dx} = \frac{v-1}{1 + v} -v\\\Rarr\space x\frac{dv}{dx} = \frac{v - 1-v-v^{2}}{1+v}$$
$$\Rarr\space -x\frac{dv}{dx} = \frac{\text{1 + v}^{2}}{\text{1 + v}}\\\Rarr\space \frac{(v+1)}{1 + v^{2}}dv = -\frac{1}{x}dx$$
On integrating both sides, we get
$$\int\frac{(v+1)}{1 + v^{2}}dv = -\int\frac{dx}{x}\\\Rarr\space\frac{v}{1 + v^{2}}dv + \int\frac{1}{1 + v^{2}}dv\\= -\int\frac{dx}{x}$$
In first integral, put
$$v^{2}+1 = t\\\Rarr\space 2v = \frac{dt}{dv}\\\Rarr\space dv = \frac{dt}{2v}\\\therefore\space\int\frac{v}{t}×\frac{dt}{2v} + \int\frac{1}{1 + v^{2}}dv \\= -\int\frac{dx}{x}\\\Rarr\space \frac{1}{2}\int\frac{1}{t}dt + \int\frac{1}{\text{1 + v}^{2}}dv = -\int\frac{dx}{x}\\\Rarr\space \frac{1}{2}\text{log}|t| + \text{tan}^{\normalsize-1}v = -\text{log}|x|+\text{C}\\\bigg[\because\space \int\frac{1}{1+x^{2}}dx = \text{tan}^{\normalsize-1}x\bigg]$$
$$\Rarr\space \frac{1}{2}\text{log}|v^{2}+1| + \text{tan}^{\normalsize-1}v\\= - \text{log}|x| + \text{C}\\\lbrack\because\space t = v^{2} +1\rbrack$$
$$\Rarr\space\frac{1}{2}\text{log}\bigg(\frac{y^{2} + x^{2}}{x^{2}}\bigg) + \text{log}|x| + \\\text{tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg)=\text{C}\\\bigg(\text{Put v} =\frac{y}{x}\bigg)\\\Rarr\space \text{log}\bigg(\frac{y^{2}+x^{2}}{x^{2}}\bigg) + 2\space\text{log}|x| +\\ 2\text{tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg) = \text{2 C}\\\Rarr\space \text{log}\bigg(\frac{y^{2}+x^{2}}{x^{2}}\bigg) + \text{log}\space x^{2} + \\2\text{tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg) = \text{2C}$$
$$\Rarr\space\text{log}\frac{(y^{2} + x^{2})x^{2}}{x^{2}} +\\ \text{2 tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg) = \text{2C}\\\Rarr\space \text{log}(x^{2} + y^{2}) + 2\text{tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg) = \text{2 C}\\\Rarr\space \text{log}(x^{2} + y^{2}) + 2\text{tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg) = A\\\lbrack\text{Put 2C = A}\rbrack\space\text{...(ii)}$$
We are given that when x = 1, y = 1
$$\therefore\space\text{log 2} + \bigg(2×\frac{\pi}{4}\bigg)= \text{A}\\\bigg(\because\space \text{tan}^{\normalsize-1}(1) = \frac{\pi}{4}\bigg)$$
On putting the value of A in eq. (ii), we get
$$\text{log}(x^{2} + y^{2}) + 2\space\text{tan}^{\normalsize-1}\bigg(\frac{y}{x}\bigg)\\=\frac{\pi}{2} + \text{log 2}$$
This is the required solution of the given differential equation.
12. x2 dy + (xy + y2) dx = 0, y = 1 when x = 1.
Sol. Given, x2dy = – (xy + y2)dx
$$\Rarr\space \frac{dy}{dx} = -\frac{xy + y^{2}}{x^{2}}$$
Thus, the given differential equation is homogeneous.
$$\text{So, put}\frac{y}{x}= v\space\text{i.e}\\ y = vx\\\Rarr\space \frac{dy}{dx} = v + x\frac{dv}{dx}\\\text{Then, eq. (i) becomes}\\v+x\frac{dv}{dx} = -\frac{x^{2}v + x^{2}v^{2}}{x^{2}}\\\Rarr\space v+x\frac{dv}{dx} = -\frac{x^{2}(v + v^{2})}{x^{2}}\\\Rarr v + x\frac{dv}{dx}=-(v + v^{2})\\\Rarr\space x\frac{dv}{dx} = -( 2v + v^{2})\\\Rarr\space \frac{dv}{v^{2} + 2v} = -\frac{1}{x}dx$$
On integrating both sides, we get
$$\Rarr\space\int\frac{dv}{v^{2} + 2v}= -\int\frac{dx}{x}\\\Rarr\space \int\frac{1}{x^{2}+ 2v+1-1}dv =- \int\frac{dx}{x}\\\Rarr\space \int\frac{1}{(v+1)^{2}-1}dv = -\int\frac{dx}{x}$$
Let v + 1 = t
$$\Rarr\space dv = dt $$
$$\therefore\space \int\frac{1}{t^{2}-1}dt = -\int\frac{dx}{x}\\\Rarr\space\frac{1}{2}\text{log}\bigg|\frac{t-1}{t+1}\bigg|=-\text{log}|x|+ \text{C}\\\bigg[\because\space \int\frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\text{log}\bigg|\frac{x-a}{x+a}\bigg| + \text{C}\bigg]\\\Rarr\frac{1}{2}\text{log}\bigg|\frac{v+1-1}{v+1+1}\bigg|\\=-\text{log}|x| + \text{log C}\\\lbrack\because\space t = v+1\rbrack\\\Rarr\space\frac{1}{2}\text{log}\bigg|\frac{v}{v+2}\bigg| + \text{log}|x| = \text{log C}\\\Rarr\space \text{log}\bigg|\frac{v}{v+2}\bigg| + 2\space\text{log}|x| = \space\text{2 log C}$$
$$\Rarr\space\text{log}\bigg|\bigg(\frac{v}{v+2}\bigg)x^{2}\bigg| = \text{log C}^{2}\\\Rarr\space \text{log}\begin{vmatrix}\bigg(\frac{\frac{y}{x}}{\frac{y}{x}+2}\bigg)x^{2}\end{vmatrix}\\=\text{log C}^{2}\bigg(\text{Put v = }\frac{y}{x}\bigg)\\\Rarr\space\frac{x^{2}y}{2x+y} = A\space\text{...(ii)}\\\lbrack\text{Let C}^{2} = A \space\text{and log m = log n}\\\Rarr\space m = n\rbrack$$
When x = 1, y = 1
$$\therefore\space\frac{1}{2 + 1} \text{= A}\\\Rarr\space A = \frac{1}{3}$$
On putting the value of A in eq. (ii), we get
$$\frac{x^{2}y}{2x+y} = \frac{1}{3}\\\Rarr\space 3x^{2}y = 2x+y$$
This is the required solution of the given differential equation.
$$\textbf{13.\space}\bigg[\textbf{x sin}^{2}\bigg(\frac{y}{x}\bigg)\textbf{-y}\bigg]\textbf{dx + x dy =0,}\\\textbf{y} = \frac{\pi}{\textbf{4}}\space\textbf{when x = 1.}\\\textbf{Sol.\space}\text{Given, sin}^{2}\bigg(\frac{y}{x}\bigg) - \frac{y}{x} + \frac{dy}{dx} = 0\\\Rarr\space \frac{dy}{dx} = \frac{y}{x} - \text{sin}^{2}\bigg(\frac{y}{x}\bigg)\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So,\space put}\space\frac{y}{x}=v\\\text{i.e\space} y=vx\Rarr\space \frac{dy}{dx} = v +x\frac{dv}{dx}\\\text{Then, eq. (i) becomes,}\\\text{v + x}\frac{dv}{dx} = v - \text{sin}^{2}v\\\Rarr\space x\frac{dv}{dx} = -\text{sin}^{2}v\\\Rarr\space \text{cosec}^{2}v \space dv = -\frac{1}{x}dx$$
On integrating both sides, we get
$$\int\text{cosec}^{2}v\space\text{dv} = -\int\frac{dx}{x}\\\Rarr\space -\text{cot v = - log}|x| + \text{C}\\\Rarr\space \text{log}|x|- \text{cot v}\bigg(\frac{y}{x}\bigg)=\text{C}\\\bigg(\text{Put v =}\frac{y}{x}\bigg)\\\text{...(ii)}\\\text{When x = 1, then y =}\frac{\pi}{4},\text{therefore,}\\\text{log}|1|\text{cot}\frac{\pi}{4} = \text{C}\\\Rarr\space \text{C = 0 - 1}=-1$$
On putting the value of C in eq. (ii), we get
$$\text{log}|x| -\text{cot}\bigg(\frac{y}{x}\bigg) =-1\\\Rarr\space \text{log}|x| - \text{cot}\bigg(\frac{y}{x}\bigg)\\= -\text{log e}\space\lbrack\because\space \text{1 = log e}\rbrack\\\Rarr\space\text{cot}\bigg(\frac{y}{x}\bigg) = \text{log}|ex|\\\lbrack\because\space \text{log m + log n = log mn}\rbrack$$
This is the required solution of the given differential equation.
$$\textbf{14.\space}\frac{\textbf{dy}}{\textbf{dx}} \textbf{-} \frac{\textbf{y}}{\textbf{x}} \textbf{+ cosec}\bigg(\frac{\textbf{y}}{\textbf{x}}\bigg) \textbf{= 0},\\\textbf{y when x=1.}\\\textbf{Sol.\space}\text{Given,\space}\frac{dy}{dx} = \frac{y}{x} - \text{cosec}\bigg(\frac{y}{x}\bigg)\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So,\space put}\space\frac{y}{x} = v\space\text{i.e.,}\\\text{y = vx}\\\Rarr\space\frac{dy}{dx} = v+x\frac{dv}{dx}\\\text{Then, equation (i) becomes,}\\\text{v+x}\frac{dv}{dx} = v-\text{cosec v}\\\Rarr\space x\frac{dv}{dx} = -\text{cosec v}\\\Rarr\space \text{sin v\space dv} = \frac{1}{x}dx$$
On integrating both sides, we get
$$\int\text{sin v\space dv} = -\int\frac{dx}{x}\\\Rarr\space\text{- cos v} = -\text{log}|x| + \text{C}\\\Rarr\space\text{log}|x| - \text{cos}\bigg(\frac{y}{x}\bigg)= \text{C}\space\text{...(ii)}$$
When x = 1, then y = 0, therefore, log 1 – cos 0 = C
$$\Rarr\space \text{C = 0 - 1}=-1$$
On putting the value of C in eq. (ii), we get
$$\text{log}|x| - \text{cos}\bigg(\frac{y}{x}\bigg)=1\\\Rarr\space \text{log}|x| + 1 = \text{cos}\frac{y}{x}\\\Rarr\space \text{log}|x| + \text{log e = cos}\bigg(\frac{y}{x}\bigg)\\\lbrack\because\space\text{log e}=1\rbrack\\\Rarr\space \text{log}|ex| = \text{cos}\bigg(\frac{y}{x}\bigg)$$
This is the required solution of the given differential equation.
$$\textbf{15.\space}\textbf{2xy + y}^{\textbf{2}}\textbf{- 2x}^{\textbf{2}}\frac{\textbf{dy}}{\textbf{dx}}\textbf{= 0},\\\textbf{y = 2 when x = 1.}\\\textbf{Sol.}\space\text{Given, 2x}^{2}\frac{dy}{dx} = 2xy + y^{2}\\\Rarr\space \frac{dy}{dx} = \frac{2xy + y^{2}}{2x^{2}}\space\text{...(i)}$$
Thus, the given differential equation is homogeneous.
$$\text{So, put}\space\frac{y}{x} = v\space\text{i.e.,\space} y = vx\\\Rarr\space\frac{dy}{dx} = v +x\frac{dv}{dx}\\\text{Then, eq. (i) becomes}\\\text{v + x}\frac{dv}{dx} = \frac{2 x^{2}v + x^{2}v^{2}}{2x^{2}}\\\Rarr\space v + x\frac{dv}{dx} = \frac{x^{2}(2v + v^{2})}{2x^{2}}\\\Rarr\space v +x\frac{dv}{dx} = v +\frac{1}{2}v^{2}\\\Rarr\space x\frac{dv}{dx} = \frac{1}{2}v^{2}\\\Rarr\space \frac{2}{v^{2}}dv = \frac{1}{x}dx$$
On integrating both sides, we get
$$2\int v^{\normalsize-2}dv = \int\frac{dx}{x}\\\Rarr\space 2\bigg(\frac{v^{-2+1}}{-2+1}\bigg) = \text{log}|x| + \text{C}\\\Rarr\space 2\bigg(\frac{v^{\normalsize-1}}{\normalsize-1}\bigg) = \text{log}|x| + \text{C}\\\Rarr\space -\frac{2}{v} = \text{log}|x| + \text{C}\\\Rarr\space -\frac{2x}{y}= \text{log}|x| + \text{C}\\\bigg(\because\space v = \frac{y}{x}\bigg)\space\text{...(ii)}$$
When x = 1, then y = 2, therefore
$$\frac{-2×1}{2} = \text{log 1 + C}\\\Rarr\space -1 = 0 +\text{C}$$
On, putting the value of C in eq. (ii), we get
$$-\frac{2x}{y} = \text{log}|x|-1\\\Rarr\space y = \frac{2}{\text{1 - log}|x|}$$
$$\textbf{16. A homogeneous equation of the form}\\\frac{\textbf{dx}}{\textbf{dy}} = \textbf{h}\bigg(\frac{\textbf{x}}{\textbf{y}}\bigg)\space\textbf{can be solved by}\\\textbf{making the subsitution.}$$
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v
Sol. (c) x = vy
Since, given equation
$$\frac{dx}{dy} = h\bigg(\frac{x}{y}\bigg)\text{is a}\\\text{homogeneous, so by the substitution}\\\frac{x}{y} = v \space\text{i.e.,}\\\text{x = vy}\\\Rarr\space\frac{dx}{dy} = v + y\frac{dv}{dy}\\\text{It becomes v +y}\frac{dv}{dy} = hv\\\Rarr\space y\frac{dv}{dy} = v(h-1)\\\Rarr\space\frac{1}{(h-1)v} dv = \frac{1}{y}dy$$
On integrating both sides, we get
$$\frac{1}{(h-1)}\int\frac{1}{v}dv = \int\frac{dy}{y}\\\Rarr\space\frac{1}{(h-1)}\text{log}|v| = \text{log}|y| + \text{C}$$
17. Which of the following is a homogeneous differential equation ?
(a) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(b) xy dx – (x3 + y3 dy = 0
(c) (x3 + 2y2) dx + 2 xy dy = 0
(d) y2 dx + (x2 – xy – y2) dy = 0
Sol. (d) y2 dx + (x2 – xy – y2) dy = 0
(a) Given equation (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 can be rewritten as
$$\frac{dy}{dx} = \frac{3y + 2x +4}{4x + 6y + 5}\\\Rarr\space \frac{dy}{dx} = \frac{x\bigg(3.\frac{y}{x} + 2 + \frac{4}{x}\bigg)}{x\bigg(4 + 6\frac{y}{x} + \frac{5}{x}\bigg)}$$
This equation is not a homogeneous.
(b) Given equations is xy dx – (x3 + y3) dy = 0
On separating the variables, we get
$$(x^{3} + y^{3})dy = xy dx\\\Rarr\space \frac{dy}{dx} = \frac{\text{xy}}{\text{x}^{3} + \text{y}^{3}}\\\Rarr\space \frac{dy}{dx} = \frac{\frac{y}{x^{2}}}{1 + \bigg(\frac{y}{x}\bigg)^{3}} $$
This equation is not a homogeneous equation.
(c) Given equation is (x3 + 2y2) dx + 2 xy dy = 0
2xy dy = – (x3 + 2y2) dx
On separating the variables, we get
$$\Rarr\space\frac{dy}{dx} = -\frac{x^{3} + 2y^{2}}{2xy}\\\Rarr\space \frac{dy}{dx} = \frac{1 + \frac{2y^{2}}{x^{3}}}{\frac{2y}{x^{2}}}$$
This equation is not homogeneous equation.
(d) Given equation y2dx + (x2 – xy – y2)dy = 0
On separating the variables, we get
(x2 – xy – y2)dy = – y2 dx
$$\Rarr\space\frac{dy}{dx} = -\frac{y^{2}}{x^{2}-xy-y^{2}}\\\Rarr\space \frac{dy}{dx} = \frac{\bigg(\frac{y}{x}\bigg)^{2}}{\bigg[1 - \frac{y}{x} - \bigg(\frac{y}{x}\bigg)^{2}\bigg]}$$
This equation is homogeneous equation.
Share page on
