NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.5

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Exercise 9.5 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 9.5

Q1. $(x^2+xy)dy=(x^2+y^2)dx$

Answer. $\begin{array}{c}\text{The given differential equation i.e.,}(x^2+xy)dy=(x^2+y^2)dx\text{can be written as:}\\ \frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}...\left(i\right)\\ \text{Let}F(x,y)=\frac{x^2+y^2}{x^2+xy}\end{array}$ $\begin{array}{c}\text{Now,}F(\lambda x,\lambda y)=\frac{(\lambda x{)}^2+(\lambda y{)}^2}{(\lambda x{)}^2+(\lambda x\left)\right(\lambda y)}=\frac{x^2+y^2}{x^2+xy}={\lambda }^0\cdot F(x,y)\\ \text{This shows that equation}\left(i\right)\text{is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \text{Differentiating both sides with respect to}x,\text{we get:}\end{array}$ $\begin{array}{c}\frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}v\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{x^2+(vx{)}^2}{x^2+x\left(vx\right)}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v=\frac{(1+v^2)-v(1+v)}{1+v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-v}{1+v}\end{array}$ $\begin{array}{c}\Rightarrow \left(\frac{1+v}{1-v}\right)=dv=\frac{dx}{x}\\ \Rightarrow \left(\frac{2-1+v}{1-v}\right)dv=\frac{dx}{x}\\ \Rightarrow (\frac{2}{1-v}-1)dv=\frac{dx}{x}\\ \text{Integrating both sides, we get:}\\ \Rightarrow v=-2\log (1-v)-\log x+\log k\\ \Rightarrow v=\log \left[\frac{k}{x(1-v{)}^2}\right]\end{array}$ $\begin{array}{c}\Rightarrow \frac{y}{x}=\log \left[\frac{k}{x{(1-\frac{y}{x})}^2}\right]\\ \Rightarrow \frac{y}{x}=\log \left[\frac{kx}{(x-y{)}^2}\right]\\ \Rightarrow \frac{kx}{(x-y{)}^2}=e^x\\ \Rightarrow (x-y{)}^2=kx{e}^{-\frac{y}{x}}\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q2. $y^{\prime }=\frac{x+y}{x}$

Answer. $\begin{array}{c}{y}^{\prime }=\frac{x+y}{x}\\ \Rightarrow \frac{dy}{dx}=\frac{x+y}{x}...\left(i\right)\\ \mathrm{Let}F(x,y)=\frac{x+y}{x}\\ \text{Now,}F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}={\lambda }^{0}F(x,y)\end{array}$ $\begin{array}{c}\text{Thus, the given equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \text{Differentiating both sides with respect to}x,\text{we get:}\\ \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{x+vx}{x}\\ \Rightarrow v+x\frac{dv}{dx}=1+v\\ x\frac{dv}{dx}=1\\ \Rightarrow dv=\frac{dx}{x}\\ \text{Integrating both sides, we get:}\\ v=logx+C\\ \Rightarrow \frac{y}{x}=\log x+C\\ \Rightarrow y=x\log x+Cx\end{array}$ This is the required solution of the given differential equation.

Q3. $(x-y)dy-(x+y)dx=0$

Answer. $\begin{array}{c}\text{The given differential equation is:}\\ (x-y)dy-(x+y)dx=0\\ \Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}...\left(i\right)\\ \text{Let}F(x,y)=\frac{x+y}{x-y}\end{array}$ $\begin{array}{c}\therefore F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}={\lambda }^0\cdot F(x,y)\\ \text{Thus, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\end{array}$ $\begin{array}{c}\Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}\end{array}$ $\begin{array}{c}x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1-v}\\ \Rightarrow \frac{1-v}{(1+v^2)}dv=\frac{dx}{x}\\ \Rightarrow (\frac{1}{1+v^2}-\frac{v}{1-v^2})dv=\frac{dx}{x}\end{array}$ $\begin{array}{c}\text{Integrating both sides, we get:}\\ {\tan }^{-1}v-\frac{1}{2}\log (1+v^2)=\log x+C\\ \Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\log [1+{\left(\frac{y}{x}\right)}^2]=\log x+C\\ \Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\log \left(\frac{x^2+y^2}{x^2}\right)=\log x+C\end{array}$ $\begin{array}{cc}& \Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}[\log (x^2+y^2)-\log x^2]=\log x+C\\ & \Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)=\frac{1}{2}\log (x^2+y^2)+C\\ & \Rightarrow \mathrm{this}\text{is the required solution of the given differential equation.}\end{array}$

Q4. $(x^2-y^2)dx+2xydy=0$

Answer. $\begin{array}{c}\text{The given differential equation is:}\\ (x^2-y^2)dx+2xydy=0\\ \Rightarrow \frac{dy}{dx}=\frac{-(x^2-y^2)}{2xy}...\left(i\right)\\ \text{Let}F(x,y)=\frac{-(x^2-y^2)}{2xy}\end{array}$ $\begin{array}{c}\therefore F(\lambda x,\lambda y)=\left[\frac{(\lambda x{)}^2-(\lambda y{)}^2}{2\left(\lambda x\right)\left(\lambda y\right)}\right]=\frac{-(x^2-y^2)}{2xy}={\lambda }^0\cdot F(x,y)\\ \text{To solve it, we given differential equation is a homogeneous equation.}\\ y=vx\end{array}$ $\begin{array}{c}\Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=-\frac{x^2-(vx{)}^2}{2x\cdot \left(vx\right)}]\\ v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\end{array}$ $\begin{array}{cc}& \Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}\\ & \Rightarrow x\frac{dv}{dx}=-\frac{(1+v^2)}{2v}\\ & \Rightarrow \frac{2v}{1+v^2}dv=-\frac{dx}{x}\\ & \text{Integrating both sides, we get:}\\ & \log (1+v^2)=-\log x+\log C=\log \frac{C}{x}\end{array}$ $\begin{array}{c}\Rightarrow 1+v^2=\frac{C}{x}\\ \Rightarrow [1+\frac{y^2}{x^2}]=\frac{C}{x}\\ \Rightarrow x^2+y^2=Cx\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q5. $x^2\frac{dy}{dx}-x^2-2y^2+xy$

Answer. $\begin{array}{c}\text{The given differential equation is:}\\ x^2\frac{dy}{dx}=x^2-2y^2+xy\end{array}$ $\begin{array}{c}\frac{dy}{dx}=\frac{x^2-2y^2+xy}{x^2}...\left(i\right)\\ \mathrm{Let}F(x,y)=\frac{x^2-2y^2+xy}{x^2}\\ \therefore F(\lambda x,\lambda y)=\frac{(\lambda x{)}^2-2(\lambda y{)}^2+\left(\lambda x\right)\left(\lambda y\right)}{(\lambda x{)}^2}=\frac{x^2-2y^2+xy}{x^2}={\lambda }^0\cdot F(x,y)\end{array}$ $\begin{array}{c}\text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{x^2-2(vx{)}^2+x\cdot (vx)}{x^2}\\ \Rightarrow v+x\frac{dv}{dx}=1-2v^2+v\\ \Rightarrow x\frac{dv}{dx}=1-2v^2\\ \Rightarrow \frac{dv}{1-2v^2}=\frac{dx}{x}\\ \Rightarrow \frac{1}{2}\cdot \frac{dv}{\frac{1}{2}-v^2}=\frac{dx}{x}\end{array}$ $\Rightarrow \frac{1}{2}\cdot \left[\frac{dv}{{\left(\frac{1}{\sqrt{2}}\right)}^2-v^2}\right]=\frac{dx}{x}$ Integrating both sides, we get : $\frac{1}{2}\cdot \frac{1}{2\times \frac{1}{\sqrt{2}}}\log \left|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\right|=\log |x|+C$ $\begin{array}{c}\Rightarrow \frac{1}{2\sqrt{2}}\log \left|\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}\right|=\log |x|+C\\ \\ \Rightarrow \frac{1}{2\sqrt{2}}\log \left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right|=\log |x|+C\end{array}$ This is the required solution for the given differential equation.

Q6. $xdy-ydx=\sqrt{x^2+y^2}dx$

Answer. $\begin{array}{c}xdy-ydx=\sqrt{x^2+y^2}dx\\ \Rightarrow xdy=[y+\sqrt{x^2+y^2}]dx\\ \frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x^2+y^2}...\left(i\right)\\ \text{Let}F(x,y)=\frac{y+\sqrt{x^2+y^2}}{x^2}\end{array}$ $\begin{array}{c}\therefore F(\lambda x,\lambda y)=\frac{\lambda x+\sqrt{(\lambda x{)}^2+(\lambda y{)}^2}}{\lambda x}=\frac{y+\sqrt{x^2+y^2}}{x}={\lambda }^0\cdot F(x,y)\\ \text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as: y=vx}\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$ $\begin{array}{c}\text{Substituting the values of}v\text{and}dx\text{in equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^2+(vx{)}^2}}{x}\\ \Rightarrow v+x\frac{dv}{dx}=v+\sqrt{1+v^2}\\ \Rightarrow \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}\\ \text{Integrating both sides, we get:}\end{array}$ $\begin{array}{c}\log |v+\sqrt{1+v^2}|=\log |x|+\log C\\ \Rightarrow \log |\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}|=\log |Cx|\\ \Rightarrow \log \left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=\log |Cx|\\ \Rightarrow y+\sqrt{x^2+y^2}=C{x}^2\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q7. $\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\}ydx=\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\}xdy$

Answer. The given differential equation is: $\begin{array}{c}\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\}ydx=\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\}xdy\\ \frac{dy}{dx}=\frac{\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\}y}{\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\}x}...\left(i\right)\\ \text{Let}F(x,y)=\frac{\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\}}{\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\}x}.\end{array}$ $\begin{array}{cc}\therefore F(\lambda x,\lambda y)& =\frac{\{\lambda x\cos \left(\frac{\lambda y}{\lambda x}\right)+\lambda y\sin \left(\frac{\lambda y}{\lambda x}\right)\}\lambda y}{{\{\lambda y\sin \left(\frac{\lambda y}{\lambda x}\right)-\lambda x\sin \left(\frac{\lambda y}{\lambda x}\right)\}}_{\lambda x}}\\ & =\frac{\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\}y}{\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\}x}\\ & ={\lambda }^0\cdot F(x,y)\end{array}$ $\begin{array}{c}\text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \Rightarrow \frac{dy}{dx}=v+x=\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation (i), we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{(x\cos v+vx\sin v)\cdot vx}{(vx\sin v-x\cos v)\cdot x}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v}{v\sin v-\cos v}\\ \Rightarrow x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v}{v\sin v-\cos v}-v\\ \Rightarrow x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v-v^2\sin v+v\cos v}{v\sin v-\cos v}\end{array}$ $\begin{array}{cc}& \Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}\\ & \Rightarrow \left[\frac{v\sin v-\cos v}{v\cos v}\right]dv=\frac{2dx}{x}\\ & \Rightarrow (\tan v-\frac{1}{v})dv=\frac{2dx}{x}\end{array}$ $\begin{array}{cc}& \text{Integrating both sides, we get:}\\ & \log \left(\sec v\right)-\log v=2\log x+\log C\\ \Rightarrow & \Rightarrow \log \left(\frac{\sec v}{v}\right)=\log \left({Cx}^2\right)\\ \Rightarrow & \Rightarrow \left(\frac{\sec v}{v}\right)=C{x}^2\\ & \Rightarrow \sec v=C{x}^{2}v\\ & \Rightarrow \sec \left(\frac{y}{x}\right)=C\cdot x^2\cdot \frac{y}{x}\end{array}$ $\begin{array}{c}\Rightarrow \sec \left(\frac{y}{x}\right)=Cxy\\ \Rightarrow \cos \left(\frac{y}{x}\right)=\frac{1}{\mathrm{Cxy}}=\frac{1}{C}\cdot \frac{1}{xy}\\ \Rightarrow xy\cos \left(\frac{y}{x}\right)=k(k=\frac{1}{c})\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q8. $x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0$

Answer. $\begin{array}{c}x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0\\ \Rightarrow x\frac{dy}{dx}=y-x\sin \left(\frac{y}{x}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{y-x\sin \left(\frac{y}{x}\right)}{x}\\ \Rightarrow \frac{dy}{dx}=\frac{y-x\sin \left(\frac{y}{x}\right)}{x}...\left(i\right)\\ \text{Let}F(x,y)=\frac{y-x\sin \left(\frac{y}{x}\right)}{x}\end{array}$ $\begin{array}{c}\therefore F(\lambda x,\lambda y)=\frac{\lambda y-\lambda x\sin \left(\frac{\lambda y}{\lambda x}\right)}{\lambda x}=\frac{y-x\sin \left(\frac{y}{x}\right)}{x}={\lambda }^0\cdot F(x,y)\\ \text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as: y=vx}\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$ $\begin{array}{c}\text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=\frac{wx-x\sin v}{x}\\ \Rightarrow v+x\frac{dv}{dx}=v-\sin v\\ \Rightarrow -\frac{dv}{\sin v}=\frac{dx}{x}\\ \Rightarrow \csc vdv=-\frac{dx}{x}\end{array}$ $\begin{array}{c}\text{Integrating both sides, we get:}\\ \log |\csc v-\cot v|=-\log x+\log C=\log \frac{C}{x}\\ \Rightarrow \csc \left(\frac{y}{x}\right)-\cot \left(\frac{y}{x}\right)=\frac{C}{x}\\ \Rightarrow \frac{1}{\sin \left(\frac{y}{x}\right)}-\frac{\cos \left(\frac{y}{x}\right)}{\sin \left(\frac{y}{x}\right)}=\frac{C}{x}\\ \Rightarrow x[1-\cos \left(\frac{y}{x}\right)]=\mathrm{Csin}\left(\frac{y}{x}\right)\end{array}$ This is the required solution of the given differential equation.

Q9. $ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$

Answer. $\begin{array}{c}ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0\\ \Rightarrow ydx=[2x-x\log \left(\frac{y}{x}\right)]dy\\ \Rightarrow \frac{dy}{dx}=\frac{y}{2x-x\log \left(\frac{y}{x}\right)}...\left(i\right)\\ \text{Let}F(x,y)=\frac{y}{2x-x\log \left(\frac{y}{x}\right)}\end{array}$ $\begin{array}{c}\therefore F(\lambda x,\lambda y)=\frac{\lambda y}{2\left(\lambda x\right)-\left(\lambda x\right)\log \left(\frac{\lambda y}{\lambda x}\right)}=\frac{y}{2x-\log \left(\frac{y}{x}\right)}={\lambda }^{\circ }\cdot F(x,y)\\ \text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\end{array}$ $\begin{array}{c}\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=\frac{vx}{2x-x\log v}\end{array}$ $\begin{array}{cc}& \Rightarrow v+x\frac{dv}{dx}=\frac{v}{2-\log v}\\ & \Rightarrow x\frac{dv}{dx}=\frac{v}{2-\log v}-v\\ & \Rightarrow x\frac{dv}{dx}=\frac{v-2v+v\log v}{2-\log v}\\ & \Rightarrow x\frac{dv}{dx}=\frac{v\log v-v}{2-\log v}\\ & \Rightarrow \frac{2-\log v}{v(\log v-1)}dv=\frac{dx}{x}\end{array}$ $\begin{array}{cc}& \Rightarrow \left[\frac{1+(1-\log v)}{v(\log v-1)}\right]dv=\frac{dx}{x}\\ & \Rightarrow [\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}\\ & \text{Integrating both sides, we get:}\\ & \int \frac{1}{v(\log v-1)}dv-\int \frac{1}{v}dv=\int \frac{1}{x}dx\\ & \Rightarrow \int \frac{dv}{v(\log v-1)}-\log v=\log x+\log C...\left(ii\right)\end{array}$ $\begin{array}{c}\Rightarrow \text{Let}\log v-1=t\\ \Rightarrow \frac{d}{dv}(\log v-1)=\frac{dt}{dv}\\ \Rightarrow \frac{1}{v}=\frac{dt}{dv}\\ \Rightarrow \frac{dv}{v}=dt\\ \text{Therefore, equation (}i\text{) becomes:}\end{array}$ $\begin{array}{c}\Rightarrow \int \frac{dt}{t}-\log v=\log x+\log C\\ \Rightarrow \log t-\log \left(\frac{y}{x}\right)=\log \left(Cx\right)\\ \Rightarrow \log [\log \left(\frac{y}{x}\right)-1]-\log \left(\frac{y}{x}\right)=\log \left(Cx\right)\\ \Rightarrow \log \left[\frac{\log \left(\frac{y}{x}\right)-1}{\frac{y}{x}}\right]=\log \left(Cx\right)\end{array}$ $\begin{array}{c}\Rightarrow \frac{x}{y}[\log \left(\frac{y}{x}\right)-1]=Cx\\ \Rightarrow \log \left(\frac{y}{x}\right)-1=Cy\\ \text{This is the required solution of the given differential equation.}\end{array}$

Answer. $\begin{array}{c}(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\\ \Rightarrow (1+e^y)dx=-e^y(1-\frac{x}{y})dy\end{array}$ $\Rightarrow \frac{dx}{dy}=\frac{-e^{\frac{r}{r}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}...\left(i\right)$ $F(x,y)=\frac{-e^{\frac{x}{r}}(1-\frac{x}{y})}{1+e^{\frac{z}{y}}}$ $\therefore F(\lambda x,\lambda y)=\frac{-e^{\frac{2x}{2y}}(1-\frac{\lambda x}{\lambda y})}{1+e^{\frac{dx}{dy}}}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{r}{y}}}={\lambda }^0\cdot F(x,y)$ $\begin{array}{c}\text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ x=vy\\ \Rightarrow \frac{d}{dy}\left(x\right)=\frac{d}{dy}\left(yy\right)\\ \Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}\\ \text{Substituting the values of}x\text{and}dy\text{in equation (i), we get:}\end{array}$ $\begin{array}{cc}v+y\frac{dv}{dy}& =\frac{-e^y(1-v)}{1+e^x}\\ \Rightarrow y\frac{dv}{dy}& =\frac{-e^r+v{e}^y}{1+e^y}-v\end{array}$ $\begin{array}{c}\Rightarrow y\frac{dv}{dy}=\frac{-e^v+v{e}^v-v-v{e}^v}{1+e^{\gamma }}\\ \Rightarrow y\frac{dv}{dy}=-\left[\frac{v+e^r}{1+e^r}\right]\end{array}$ $\begin{array}{c}\Rightarrow \left[\frac{1+e^v}{v+e^y}\right]dv=-\frac{dy}{y}\\ \text{Integrating both sides, we get:}\end{array}$ $\begin{array}{cc}& \Rightarrow \log (v+e^v)=-\log y+\log C=\log \left(\frac{C}{y}\right)\\ & \Rightarrow [\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}\\ & \Rightarrow x+y{e}^{\frac{x}{y}}=C\\ & \text{This is the required solution of the given differential equation.}\end{array}$

Q11. $(x+y)dy+(x-y)dy=0;y=1\text{when}x=1$

Answer. $\begin{array}{c}(x+y)dy+(x-y)dx=0\\ \Rightarrow (x+y)dy=-(x-y)dx\\ \Rightarrow \frac{dy}{dx}=\frac{-(x-y)}{x+y}...\left(i\right)\\ \text{Let}F(x,y)=\frac{-(x-y)}{x+y}\\ \therefore F(\lambda x,\lambda y)=\frac{-(\lambda x-\lambda y)}{\lambda x-\lambda y}=\frac{-(x-y)}{x+y}={\lambda }^0\cdot F(x,y)\end{array}$ $\begin{array}{c}\text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}dx\text{in equation (i), we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{-(x-vx)}{x+vx}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{v-1}{v+1}\\ \Rightarrow x\frac{dv}{dx}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1}\\ \Rightarrow x\frac{dv}{dx}=\frac{v-1-v^2-v}{v+1}=\frac{-(1+v^2)}{v+1}\\ \Rightarrow \frac{(v+1)}{1+v^2}dv=-\frac{dx}{x}\end{array}$ $\begin{array}{cc}& \text{Integrating both sides, we get:}\\ & \frac{1}{2}\log (1+v^2)+{\tan }^{-1}v=-\log x+k\\ \Rightarrow & \Rightarrow \log (1+v^2)+2{\tan }^{-1}v=-2\log x+2k\\ & \Rightarrow \log [(1+v^2)\cdot x^2]+2{\tan }^{-1}v=2k\\ & \Rightarrow \log [(1+\frac{y^2}{x^2})\cdot x^2]+2{\tan }^{-1}\frac{y}{x}=2k\\ & \Rightarrow \log (x^2+y^2)+2{\tan }^{-1}\frac{y}{x}=2k...\left(ii\right)\end{array}$ $\begin{array}{c}\text{Now,}y=1\text{at}x=1\\ \Rightarrow \log 2+2{\tan }^{-1}1=2k\\ \Rightarrow \log 2+2\times \frac{\pi }{4}=2k\\ \Rightarrow \frac{\pi }{2}+\log 2=2k\\ \text{Substituting the value of}2k\text{in equation}\left(ii\right),\text{we get:}\\ \log (x^2+y^2)+2{\tan }^{-1}\left(\frac{y}{x}\right)=\frac{\pi }{2}+\log 2\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q12. $x^{2}dy+(xy+y^2)dx=0;y=1\text{when}x=1$

Answer. $\begin{array}{c}{x}^{2}dy+(xy+y^2)dx=0\\ \Rightarrow x^{2}dy=-(xy+y^2)dx\\ \Rightarrow \frac{dy}{dx}=\frac{-(xy+y^2)}{x^2}...\left(i\right)\\ \text{Let}F(x,y)=\frac{-(xy+y^2)}{x^2}\\ \therefore F(\lambda x,\lambda y)=\frac{[\lambda x\cdot \lambda y+(\lambda y{)}^2]}{(\lambda x{)}^2}=\frac{-(xy+y^2)}{x^2}={\lambda }^0\cdot F(x,y)\end{array}$ $\begin{array}{c}\text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}dx\text{in equation}\left(i\right),\text{we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{-[x\cdot vx+(vx{)}^2]}{x^2}=-v-v^2\\ \Rightarrow x\frac{dv}{dx}=-v^2-2v=-v(v+2)\\ \Rightarrow \frac{dv}{v(v+2)}=-\frac{dx}{x}\\ \Rightarrow \frac{1}{2}\left[\frac{(v+2)-v}{v(v+2)}\right]dv=-\frac{dx}{x}\\ \Rightarrow \frac{1}{2}[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}\end{array}$ $\begin{array}{c}\text{Integrating both sides, we get:}\\ \frac{1}{2}[\log v-\log (v+2\left)\right]=-\log x+\log C\\ \Rightarrow \frac{1}{2}\log \left(\frac{v}{v+2}\right)=\log \frac{C}{x}\\ \Rightarrow \frac{v}{v+2}={\left(\frac{C}{x}\right)}^2\end{array}$ $\begin{array}{cc}& \Rightarrow \frac{\frac{y}{x}}{\frac{y}{x}+2}={\left(\frac{C}{x}\right)}^2\\ & \Rightarrow \frac{y}{y+2x}=\frac{C^2}{x^2}\\ \Rightarrow & \frac{x^{2}y}{y+2x}=C^2...\left(ii\right)\\ \text{Now,}y& =1\text{at}x=1\end{array}$ $\begin{array}{c}\Rightarrow \frac{1}{1+2}=C^2\\ \Rightarrow C^2=\frac{1}{3}\end{array}$ Substituting $\begin{array}{c}{C}^2=\frac{1}{3}\text{in equation}\left(ii\right),\text{we get:}\\ \frac{x^{2}y}{y+2x}=\frac{1}{3}\\ \Rightarrow y+2x=3x^{2}y\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q13. $\left[x{\sin }^2(\frac{x}{y}-y)\right]dx+xdy=0;y\frac{\pi }{4}\text{when}x=1$

Answer. $\begin{array}{c}[x{\sin }^2\left(\frac{y}{x}\right)-y]dx+xdy=0\\ \Rightarrow \frac{dy}{dx}=\frac{-[x{\sin }^2\left(\frac{y}{x}\right)-y]}{x}\\ \mathrm{Let}F(x,y)=\frac{-[x{\sin }^2\left(\frac{y}{x}\right)-y]}{x}\end{array}$ $\begin{array}{c}\therefore F(\lambda x,\lambda y)=\frac{-[\lambda x\cdot {\sin }^2\left(\frac{\lambda x}{\lambda y}\right)-\lambda y]}{\lambda x}=\frac{-[x{\sin }^2\left(\frac{y}{x}\right)-y]}{x}={\lambda }^0\cdot F(x,y)\\ \text{To solve this diven differential equation is a homogeneous equation.}\\ y=vx\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x=\frac{dv}{dx}\end{array}$ $\begin{array}{c}\text{Substituting the values of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=\frac{-[x{\sin }^{2}v-vx]}{x}\\ \Rightarrow v+x\frac{dv}{dx}=-[{\sin }^{2}v-v]=v-{\sin }^{2}v\\ \Rightarrow x\frac{dv}{dx}=-{\sin }^{2}v\\ \Rightarrow \frac{dv}{{\sin }^{2}v}=-\frac{dx}{dx}\\ \Rightarrow {\csc }^{2}vdv=-\frac{dx}{x}\end{array}$ Integrating both sides, we get: $\begin{array}{c}-\cot v=-\log |x|-C\\ \Rightarrow \cot v=\log |x|+C\\ \Rightarrow \cot \left(\frac{y}{x}\right)=\log |x|+\log C\\ \Rightarrow \cot \left(\frac{y}{x}\right)=\log |Cx|...\left(ii\right)\\ now,y=\frac{\pi }{4}\text{at}x=1\end{array}$ $\begin{array}{c}\Rightarrow \cot \left(\frac{\pi }{4}\right)=\log |C|\\ \Rightarrow 1=\log C\\ \Rightarrow C=e^1=e\\ \text{Substituting}C=e\text{in equation}\left(ii\right),\text{we get:}\\ \cot \left(\frac{y}{x}\right)=\log |ex|\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q14. $\frac{dy}{dx}-\frac{y}{x}+\csc \left(\frac{y}{x}\right)=0;y=0\text{when}x=1$

Answer. $\begin{array}{c}\frac{dy}{dx}-\frac{y}{x}+\csc \left(\frac{y}{x}\right)=0\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\csc \left(\frac{y}{x}\right)...\left(i\right)\\ \mathrm{Let}F(x,y)=\frac{y}{x}-\csc \left(\frac{y}{x}\right)\\ \therefore F(\lambda x,\lambda y)=\frac{y}{\lambda x}-\csc \left(\frac{y}{x}\right)\\ \Rightarrow F(\lambda x,\lambda y)=\frac{y}{x}-\csc \left(\frac{y}{x}\right)=F(x,y)={\lambda }^0\cdot F(x,y)\end{array}$ Therefore, the given differential equation is a homogeneous equation. $\begin{array}{c}\text{To solve it, we make the substitution as:}\\ y=vx\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the values of}y\text{and}\overline{dx}\text{in equation}\left(i\right),\text{we get:}\\ v+x\frac{dv}{dx}=v-\csc v\end{array}$ $\begin{array}{c}\Rightarrow -\frac{dv}{\csc v}=-\frac{dx}{x}\\ \Rightarrow -\sin vdv=\frac{dx}{x}\\ \text{Integrating both sides, we get:}\\ \cos v=\log x+\log C=\log |Cx|\\ \Rightarrow \cos \left(\frac{y}{x}\right)=\log |Cx|...\left(ii\right)\end{array}$ $\begin{array}{c}\text{This is the required solution of the given differential equation.}\\ \text{Now,}y=0\text{at}x=1\\ \Rightarrow \cos \left(0\right)=\log C\\ \Rightarrow 1=\log C\\ \Rightarrow C=e^1=e\\ \text{Substituting}C=e\text{in equation}\left(ii\right),\text{we get:}\end{array}$ $\begin{array}{c}\cos \left(\frac{y}{x}\right)=\log \left(ex\right)\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q15. $2xy+y^2-2x^2\frac{dy}{dx}=0;y=2\text{when}x=1$

Answer. $\begin{array}{c}2xy+y^2-2x^2\frac{dy}{dx}=0\\ \Rightarrow 2x^2\frac{dy}{dx}=2xy+y^2\\ \Rightarrow \frac{dy}{dx}=\frac{2xy+y^2}{2x^2}...\left(i\right)\\ \text{Let}F(x,y)=\frac{2xy+y^2}{2x^2}\\ \therefore F(\lambda x,\lambda y)=\frac{2\left(\lambda x\right)\left(\lambda y\right)+(\lambda y{)}^2}{2(\lambda x{)}^2}=\frac{2xy+y^2}{2x^2}={\lambda }^0\cdot F(x,y)\end{array}$ $\begin{array}{c}\text{Therefore, the given differential equation is a homogeneous equation.}\\ \text{To solve it, we make the substitution as:}\\ y=vx\\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \text{Substituting the value of}y\text{and}\frac{dy}{dx}\text{in equation}\left(i\right),\text{we get:}\end{array}$ $\begin{array}{c}v+x\frac{dv}{dx}=\frac{2x\left(vx\right)+(vx{)}^2}{2x^2}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{2v+v^2}{2}\\ \Rightarrow v+x\frac{dv}{dx}=v+\frac{v^2}{2}\\ \Rightarrow \frac{2}{v^2}dv=\frac{dx}{x}\\ \text{Integrating both sides, we get:}\end{array}$ $\begin{array}{cc}& 2\cdot \frac{v^{-2+1}}{-2+1}=\log |x|+C\\ & \Rightarrow -\frac{2}{v}=\log |x|+C\\ & \Rightarrow -\frac{2}{y}=\log |x|+C\\ & \Rightarrow -\frac{2x}{y}=\log |x|+C...\left(ii\right)\end{array}$ $\begin{array}{c}\text{Now,}y=2\text{at}x=1\\ \Rightarrow -1=\log \left(1\right)+C\\ \Rightarrow C=-1\\ \text{Substituting}C=-1\text{in equation}\left(ii\right),\text{we get:}\\ -\frac{2x}{y}=\log |x|-1\\ \Rightarrow \frac{2x}{y}=1-\log |x|\end{array}$ $\begin{array}{c}\Rightarrow y=\frac{2x}{1-\log |x|},(x\ne 0,x\ne e)\\ \text{This is the required solution of the given differential equation.}\end{array}$

Q16. $\begin{array}{c}\text{A homogeneous differential equation of the form}\frac{dx}{dy}=h\left(\frac{x}{y}\right)\text{can be solved by making the}\\ \text{substitution}\\ \text{A.}y=vx\\ \text{B.}v=yx\\ \text{C.}x=vy\\ \text{D.}x=v\end{array}$

Answer. For solving the homogeneous equation of the form $\frac{dx}{dy}=h\left(\frac{x}{y}\right)$, we need to make the substitution as x = vy. Hence, the correct answer is C.

Q17. $\begin{array}{c}\text{Which of the following is a homogeneous differential equation?}\\ \text{A.}(4x+6y+5)dy-(3y+2x+4)dx=0\\ \text{B.}\left(xy\right)dx-(x^3+y^3)dy=0\\ \text{C.}(x^3+2y^2)dx+2xydy=0\\ \text{D.}{y}^{2}dx+(x^2-x{y}^2-y^2)dy=0\end{array}$

Answer. $\begin{array}{c}\text{Function}F(x,y)\text{is said to be the homogenous function of degree}n,\text{if}\\ F(\lambda x,\lambda y)={\lambda }^{n}F(x,y)\text{for any non-zero constant}\left(\lambda \right)\text{.}\\ \text{Consider the equation given in alternativeD:}\\ y^{2}dx+(x^2-xy-y^2)dy=0\\ \Rightarrow \frac{dy}{dx}=\frac{-y^2}{x^2-xy-y^2}=\frac{y^2}{y^2+xy-x^2}\end{array}$ $\begin{array}{c}\\ \text{Let}F(x,y)=\frac{y^2}{y^2+xy-x^2}\\ \Rightarrow F(\lambda x,\lambda y)=\frac{(\lambda y{)}^2}{(\lambda y{)}^2+(\lambda x\left)\right(\lambda y)-(\lambda x{)}^2}\\ =\frac{{\lambda }^2{y}^2}{{\lambda }^2(y^2+xy-x^2)}\\ ={\lambda }^0\left(\frac{y^2}{y^2+xy-x^2}\right)\\ ={\lambda }^0\left(\frac{y^2}{y^2+xy-x^2}\right)\\ ={\lambda }^0\cdot F(x,y)\\ \text{Hence, the differential equation given in alternative D is a homogenous equation.}\end{array}$

Chapter-9 (Determinants)

• Exercise 9.1
• Exercise 9.2
• Exercise 9.3
• Exercise 9.4
• Exercise 9.6