Question

Solve each of the following initial value problems:
(i) (x² + y²) dx = 2xy dy, y (1) = 0

(ii) $x{e}^{y/x}-y+x\frac{dy}{dx}=0,y\left(e\right)=0$

(iii) $\frac{dy}{dx}-\frac{y}{x}+\mathrm{cosec}\frac{y}{x}=0,y\left(1\right)=0$

(iv) (xy − y²) dx − x² dy = 0, y(1) = 1

(v) $\frac{dy}{dx}=\frac{y\left(x+2y\right)}{x\left(2x+y\right)},y\left(1\right)=2$

(vi) (y⁴ − 2x³ y) dx + (x⁴ − 2xy³) dy = 0, y (1) = 1

(vii) x (x² + 3y²) dx + y (y² + 3x²) dy = 0, y (1) = 1

(viii) $\left\{x{\sin }^2\left(\frac{y}{x}\right)-y\right\}dx+xdy=0,y\left(1\right)=\frac{\mathrm{\pi }}{4}$

(ix) $x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0,y\left(2\right)=x$

Answer 1

(i) (x² + y²)dx = 2xy dy, y(1) = 0
We have,
(x² + y²) dx = 2xy .....(i)

This is a homogenous equation, so let us take y = vx

$$\begin{gathered} \mathrm{Then},\frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Putting}y=vx\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right) \\ \left(x^2+v^2{x}^2\right)=2v{x}^2\left(v+x\frac{dv}{dx}\right) \\ {x}^2\left(1+v^2\right)=2v{x}^2\left(v+x\frac{dv}{dx}\right) \\ \left(1+v^2\right)=2v^2+2vx\frac{dv}{dx} \\ 1-v^2=2vx\frac{dv}{dx} \\ \frac{dx}{x}=\frac{2vdv}{1-v^2} \end{gathered}$$
On integrating both sides, we get
$$\begin{gathered} \int \frac{1}{x}dx=\int \frac{2v}{1-v^2}dv \\ \mathrm{Let},\left(1-v^2\right)=t \\ \Rightarrow -2vdv=dt \\ {\log }_{e}x=-\int \frac{dt}{t} \\ {\log }_{e}x=-{\log }_{e}t+c \\ {\log }_{e}x=-{\log }_e\left(1-\frac{y^2}{x^2}\right)+c \\ {\log }_e\left[x\left(\frac{x^2-y^2}{x^2}\right)\right]=c \\ {\log }_e\left(\frac{x^2-y^2}{x}\right)=c \\ \mathrm{As}y\left(1\right)=0 \\ \Rightarrow c=0 \\ \therefore {\log }_e\left(\frac{x^2-y^2}{x}\right)=0 \\ \Rightarrow \frac{x^2-y^2}{x}=1 \\ \Rightarrow x^2-y^2=x \end{gathered}$$

(ii) $x{e}^{\frac{y}{x}}-y+x\frac{dy}{dx}=0y\left(e\right)=0$
This is also a homogenous equation,

Put y = vx

$$\begin{gathered} \frac{dy}{dx}=v+x\frac{dv}{dx} \\ x{e}^v-vx+x\left(v+x\frac{dv}{dx}\right)=0 \\ x{e}^v-vx+xv+x^2\frac{dv}{dx}=0 \\ x{e}^v+x^2\frac{dv}{dx}=0 \\ {e}^v=-x\frac{dv}{dx} \\ \frac{dx}{x}=-\frac{1}{e^v}dv \end{gathered}$$
On integration both sides we get,
$$\begin{gathered} \int \frac{dx}{x}=-\int \frac{1}{e^v}dv \\ {\log }_{e}x=-\int e^{-v}dv \\ \Rightarrow {\log }_{e}x=e^{-\frac{y}{x}}+c\left(\because y=vx\right) \\ \mathrm{As}\mathrm{given}y\left(e\right)=0 \\ {\log }_{e}e=e^{-\frac{0}{e}}+c \\ 1=1+c \\ \Rightarrow c=0 \\ \therefore {\log }_{e}x=e^{-\frac{y}{x}} \end{gathered}$$

(iii) $\frac{dy}{dx}-\frac{y}{x}+\mathrm{cosec}\frac{y}{x}=0,y\left(1\right)=0$
This is an homogenous equation, put y = vx
$$\begin{gathered} \frac{dy}{dx}+v+x\frac{dv}{dx} \\ v+x\frac{dv}{dx}-v+\mathrm{cosec}v=0 \\ x\frac{dv}{dx}=\mathrm{cosec}v \\ \frac{dv}{\mathrm{cosec}v}=\frac{dx}{x} \\ \sin vdv=\frac{dx}{x} \end{gathered}$$
On integrating both sides, we get
$$\begin{gathered} \int \sin vdv=\int \frac{dx}{x} \\ -\cos v={\log }_{e}x+c \\ -\cos v+{\log }_{e}x=c \\ \cos v+{\log }_{e}x=-c \\ \cos \left(\frac{y}{x}\right)+{\log }_{e}x=-c \\ Asy\left(1\right)=0 \\ \cos \left(\frac{0}{1}\right)=0+{\log }_{e}1=-c \\ 1+0=-c \\ \Rightarrow c=-1 \\ \Rightarrow \cos \left(\frac{y}{x}\right)+{\log }_{e}x=1 \end{gathered}$$

(iv) (xy − y²) dx − x² dy = 0, y(1) = 1
This is an homogenous equation, put y = vx

$$\begin{gathered} \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \left(xy-y^2\right)=x^2\left(\frac{dy}{dx}\right) \\ \left(v{x}^2-v^2{x}^2\right)=x^2\left(v+x\frac{dv}{dx}\right) \\ v{x}^2\left(1-v\right)=x^2\left(v+x\frac{dv}{dx}\right) \\ v\left(1-v\right)=v+x\frac{dv}{dx} \\ v-v^2=v+x\frac{dv}{dx} \\ -v^2=x\frac{dv}{dx} \\ -\frac{1}{x}dx=\frac{1}{v^2}dv \end{gathered}$$
On integrating both sides we get,
$$\begin{gathered} -\int \frac{1}{x}dx=\int \frac{1}{v^2}dv \\ -{\log }_{e}x=\frac{v^{-2+1}}{-2+1}+c \\ -{\log }_{e}x=\frac{v^{-1}}{-1}+c \\ -{\log }_{e}x=-\frac{1}{v}+c \\ -{\log }_{e}x=-\frac{1}{v}+c \\ \frac{x}{y}-{\log }_{e}x=c \\ Asy\left(1\right)=1 \\ \frac{1}{1}-{\log }_{e}1=c \\ \Rightarrow c=1 \end{gathered}$$

(v) $\frac{dy}{dx}=\frac{y\left(x+2y\right)}{x\left(2x+y\right)},y\left(1\right)=1$
This is an homogenous equation, put y = vx
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$$\begin{gathered} \Rightarrow v+x\frac{dv}{dx}=\frac{v\left(x+2vx\right)}{\left(2x+vx\right)} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{v\left(1+2v\right)}{\left(2+v\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{v\left(1+2v\right)-v\left(2+v\right)}{\left(2+v\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{v+2v^2-2v-v^2}{\left(2+v\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{v^2-v}{\left(2+v\right)} \\ \Rightarrow \frac{\left(2+v\right)dv}{\left(v^2-v\right)}=\frac{dx}{x} \end{gathered}$$
On integrating both side of the equation we get,
$$\begin{gathered} \int \frac{2+v}{\left(v^2-v\right)}dv=\int \frac{dx}{x} \\ \Rightarrow \int \frac{2}{v\left(v-1\right)}dv+\int \frac{v}{v\left(v-1\right)}dv=\int \frac{dx}{x} \\ \Rightarrow 2\left[\int \frac{1}{\left(1-v\right)}dv-\int \frac{1}{v}dv\right]+\int \frac{1}{v-1}dv={\log }_{e}x+c \\ \Rightarrow 2\left[{\log }_e\left(v-1\right)-{\log }_{e}v\right]+{\log }_e\left(v-1\right)={\log }_{e}x+c \\ 2\left[{\log }_e\left(\frac{v-1}{v}\right)\right]+{\log }_e\left(v-1\right)={\log }_{e}x+c \\ 2{\log }_e\left(\frac{y-x}{y}\right)+{\log }_e\left(\frac{y-x}{x}\right)={\log }_{e}x+c \\ Asy\left(1\right)=2 \\ 2{\log }_e\left(\frac{2-1}{2}\right)+{\log }_e\left(\frac{2-1}{1}\right)={\log }_{e}1+c \\ 2{\log }_e\frac{1}{2}+{\log }_{e}1={\log }_{e}1+c \\ -2{\log }_{e}2+0=0+c \\ -2{\log }_{e}2=c \\ \therefore 2{\log }_e\left(\frac{y-x}{y}\right)+{\log }_e\left(\frac{y-x}{x}\right)={\log }_{e}x-2{\log }_{e}2 \end{gathered}$$

(vi) (y⁴ − 2x³ y) dx + (x⁴ − 2xy³) dy = 0, y (1) = 1
This is an homogenous equation, put y= vx
$$\begin{gathered} \left(v^4{x}^4-2v{x}^4\right)+\left(x^4-2v^3{x}^4\right)\left[v+x\frac{dv}{dx}\right]=0 \\ \left(v^4{x}^4-2v{x}^4\right)=\left(2v^3{x}^4-x^4\right)\left[v+x\frac{dv}{dx}\right] \\ v{x}^4\left(v^3-2\right)=x^4\left(2v^3-1\right)\left[v+x\frac{dv}{dx}\right] \\ v\left(v^3-2\right)=\left(2v^3-1\right)v+x\left(2v^3-1\right)\frac{dv}{dx} \\ v\left[v^3-2-2v^3+1\right]=x\left(2v^3-1\right)\frac{dv}{dx} \\ v\left(-1-v^3\right)=x\left(2v^3-1\right)\frac{dv}{dx} \\ v\left(1+v^3\right)=x\left(1-2v^3\right)\frac{dv}{dx} \\ \frac{dx}{x}=\frac{\left(1-2v^3\right)}{v\left(1+v^3\right)}dv \end{gathered}$$
On integrating both side of the equation we get,
$$\begin{gathered} \int \frac{dx}{x}=\int \frac{\left(1-2v^3\right)}{v\left(1+v^3\right)}dv \\ \Rightarrow {\log }_{e}x=\int \frac{1+v^3-3v^3}{v\left(1+v^3\right)}dv \\ \Rightarrow {\log }_{e}x=\int \frac{1+v^3}{v\left(1+v^3\right)}dv-\int \frac{3v}{v\left(1+v^3\right)}dv \\ \Rightarrow {\log }_{e}x=\int \frac{1}{v}dv-\int \frac{3v^2}{\left(1+v^3\right)}dv \\ \Rightarrow {\log }_{e}x={\log }_{e}v-\int \frac{dt}{t} \\ \Rightarrow {\log }_{e}x={\log }_{e}v-{\log }_e\left(1+v^3\right)+clet\left(1+v^3\right)=t,3v^{2}dv=dt \\ \Rightarrow {\log }_{e}x={\log }_e\frac{v}{1+v^3}+c \\ Asv=\frac{y}{x} \\ \Rightarrow {\log }_{e}x={\log }_e\frac{{\displaystyle \frac{y}{x}}}{1+y^{{\displaystyle \frac{3}{x}}}}+c \\ \Rightarrow {\log }_{e}x={\log }_e\frac{y{x}^2}{x^3+y^3}+c \\ \mathrm{As}y\left(1\right)=1 \\ \Rightarrow {\log }_{e}1={\log }_e\frac{1}{1+1}+c \\ \Rightarrow 0={\log }_e\frac{1}{2}+c \\ c=-{\log }_e\frac{1}{2} \\ \Rightarrow c={\log }_{e}2 \\ \therefore {\log }_{e}x={\log }_e\frac{y{x}^2}{x^3+y^3}+{\log }_{e}2 \end{gathered}$$

$$\begin{gathered} \left(vii\right)x(x^2+3y^2)dx+y(y^2+3x^2)dy=0,y\left(1\right)=1 \\ \frac{dy}{dx}=\frac{-x(x^2+3y^2)}{y(y^2+3x^2)} \\ \mathrm{it}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{equation}.\mathrm{Put}y=vx \\ and\frac{dy}{dx}=v+x\frac{dv}{dx} \end{gathered}$$

$$\begin{gathered} \mathrm{So},v+x\frac{dv}{dx}=-\frac{x(x^2+3v^2{x}^2)}{vx(v^2{x}^2+3x^2)} \\ x\frac{dv}{dx}=-\frac{(1+3v^2)}{v(v^2+3)}-3 \\ =\frac{-1-3v^2-v^4-3v^2}{v(v^2+3)} \\ x\frac{dv}{dx}=\frac{-v^4-6v^2-1}{v(v^2+3)} \\ \frac{v(v^2+3)}{v^4+6v^2+1}dv=-\frac{dx}{x} \\ \int \frac{4v^3+12v}{v^4+6v^2+1}dv=-4\int \frac{dx}{x} \\ \log \left|v^4+6v^2+1\right|=\log \left|\frac{c}{x^4}\right| \\ \left|v^4+6v^2+1\right|=\left|\frac{c}{x^4}\right| \\ \left|y^4+6y^2{x}^2+x^4\right|=\left|c\right|....\left(1\right) \end{gathered}$$

$$\begin{gathered} puty=1,x=1 \\ (1+6+1)=c\Rightarrow c=8 \\ putc=8inequation\left(1\right), \\ (y^4+x^4+6x^2{y}^2)=8 \end{gathered}$$

$$\begin{gathered} \left(viii\right)\{x{\sin }^2\left(\frac{y}{x}\right)-y\}dx+xdy=0,y\left(1\right)=\frac{\mathrm{\pi }}{4} \\ \mathrm{it}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{equation}.\mathrm{so},\mathrm{we}\mathrm{put}y=vx \\ \frac{dy}{dx}=v+x\frac{dv}{dx} \\ so,v+x\frac{dv}{dx}=-{\sin }^2\left(\frac{vx}{x}\right)+\frac{vx}{x} \\ x\frac{dv}{dx}=-si{n}^{2}v \\ \frac{dv}{{\sin }^{2}v}=-\frac{dx}{x} \\ \mathrm{integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ cot\left(\frac{y}{x}\right)=\log \left|cx\right| \end{gathered}$$
$$\begin{gathered} \mathrm{putting}\mathrm{the}\mathrm{values}\mathrm{of}x=1andy=\frac{\mathrm{\pi }}{4} \\ cot\left(\frac{\mathrm{\pi }}{4}\right)=\log c \\ 1=\log c \\ c=e \\ Hence,cot\left(\frac{y}{x}\right)=\log \left(ex\right) \end{gathered}$$

$$\begin{gathered} \left(ix\right)x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0,y\left(2\right)=\mathrm{\pi } \\ \mathrm{it}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{equation}.\mathrm{put}y\mathit{=}vx \\ \mathrm{and}\mathit{}\frac{dy}{dx}\mathit{=}v\mathit{+}x\frac{dv}{dx} \\ so\mathit{,}\mathit{}v\mathit{+}x\frac{dv}{dx}\mathit{=}\frac{vx}{x}\mathit{-}\mathit{sin}\left(\frac{vx}{x}\right) \\ x\frac{dv}{dx}\mathit{=}\mathit{-}\mathit{sin}v \\ \frac{dv}{\mathit{sin}v}\mathit{=}\mathit{-}\frac{dx}{x} \\ \mathit{cos}ec\mathit{\left(}v\mathit{\right)}dv\mathit{=}\mathit{-}\frac{dx}{x} \\ \mathrm{integraing}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \mathit{log}\mathit{\left(}\mathit{cos}ec\mathit{\right(}v\mathit{)}\mathit{-}cot\mathit{(}v\mathit{\left)}\mathit{\right)}\mathit{=}\mathit{-}\mathit{log}\mathit{}x\mathit{}\mathit{+}\mathit{}\mathit{log}\mathit{}c \end{gathered}$$
$$\begin{gathered} log\left(\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\left(\frac{\mathit{y}}{\mathit{x}}\right)\mathit{-}\mathit{c}\mathit{o}\mathit{t}\left(\frac{\mathit{y}}{\mathit{x}}\right)\right)\mathit{=}\mathit{-}log\mathit{}x\mathit{}\mathit{+}\mathit{}log\mathit{}c \\ putting\mathit{}the\mathit{}values\mathit{}x\mathit{=}\mathit{2}\mathit{}and\mathit{}y\mathit{=}\mathrm{\pi }\mathit{} \\ log\left(\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\left(\frac{\mathit{\pi }}{\mathit{2}}\right)\mathit{-}\mathit{c}\mathit{o}\mathit{t}\left(\frac{\mathit{\pi }}{\mathit{2}}\right)\right)\mathit{=}\mathit{-}log\mathit{}\mathit{2}\mathit{}\mathit{+}\mathit{}log\mathit{}c \\ c\mathit{=}\mathit{0} \\ log\left(\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\left(\frac{\mathit{y}}{\mathit{x}}\right)\mathit{-}\mathit{c}\mathit{o}\mathit{t}\left(\frac{\mathit{y}}{\mathit{x}}\right)\right)\mathit{=}\mathit{-}log\mathit{}x\mathit{} \end{gathered}$$