Question

Find the particular solution of the differential equation $x^{2}dy=(2xy+y^2)dx,$ given that y= 1 when x = 1.

OR

Find the particular solution of the differential equation $(1+x^2)\frac{dy}{dx}=(e^{mta{n}^{-1}x}-y),$ given that y =1 when x = 0.

Answer 1

$x^{2}dy=(2xy+y^2)dx$
$\Rightarrow \frac{dy}{dx}=\frac{2xy+y^2}{x^2}......\left(i\right)$
$Lety=vx,$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
$\text{Substituting in (i), we get}$
$v+x\frac{dv}{dx}=\frac{2v{x}^2+v^2{x}^2}{x^2}$
$\Rightarrow v+x\frac{dv}{dx}=2v+v^2$
$\Rightarrow x\frac{dv}{dx}=v+v^2$
$\Rightarrow \frac{dv}{v^2+v}=\frac{dx}{x}$

Integrating both sides.

$\int \frac{dv}{v^2+v}=\int \frac{dx}{x}$
$\Rightarrow \int \frac{v+1-v}{v(v+1)}.dv=\int \frac{dx}{x}$

$\Rightarrow \int \frac{dv}{v}-\int \frac{dv}{v+1}=\int \frac{dx}{x}$
$\Rightarrow logv-log|v+1|=logx+logC$
$\Rightarrow log|\frac{v}{v+1}|=log|Cx|$
$\Rightarrow log|\frac{\frac{y}{x}}{\frac{y}{x}+1}|=log|Cx|$
$\Rightarrow \frac{y}{y+x}=Cx\text{[Removing logarithm in both sides]}$
$\therefore y=Cxy+C{x}^2,\text{which is the general solution}.$
$\text{Putting y=1 and x=1},$
$1=C+C$
$\Rightarrow 2C=1$
$\Rightarrow C=\frac{1}{2}$
$y=\frac{xy}{2}+\frac{x^2}{2}$
$\therefore 2y=xy+x^2,\text{which is the particular solution.}$
OR
$(1+x^2)\frac{dy}{dx}=e^{mta{n}^{-1x}}-y$
$\Rightarrow \frac{dy}{dx}=\frac{e^{mta{n}^{-1x}}}{(1+x^2)}-\frac{y}{(1+x^2)}$
$\Rightarrow \frac{dy}{dx}+\frac{y}{(1+x^2)}=\frac{e^{mta{n}^{-1x}}}{(1+x^2)}$

$P=\frac{1}{(1+x^2)},Q=\frac{e^{mta{n}^{-1}x}}{(1+x^2)}I.F.=e^{\int Pdx}=e^{\int \frac{1}{(1+x^2)}dx}=e^{ta{n}^{-1}x}\text{Thus the solution is}y{e}^{\int Pdx}=\int Q{e}^{\int Pdx}dx\Rightarrow y\times e^{ta{n}^{-1}x}=\int \frac{e^{mta{n}^{-1}x}}{(1+x^2)}.e^{ta{n}^{-1}x}.dx\Rightarrow y\times e^{ta{n}^{-1}x}=\int \frac{e^{(m+1)ta{n}^{-1}x}}{(1+x^2)}.dx\left(i\right)\int \frac{e(m+)ta{n}^{-1}x}{(1+x^2)}dx.......\left(ii\right)Let(m+1)ta{n}^{-1}x=z\frac{(m+1)}{(1+x^2)}dx=dz\frac{dx}{(1+x^2)}=\frac{dz}{(m+1)}\text{Substituting in (ii)},\frac{1}{(m+1)}\int e^{z}dz=\frac{e^z}{(m+1)}=\frac{e^{(m+1)ta{n}^{-1}x}}{(m+1)}\text{Substituting in (i)},\Rightarrow y\times e^{ta{n}^{-1}x}=\frac{e^{(m+1)ta{n}^{-1}x}}{(m+1)}+C........\left(iii\right)\text{Putting y=1 and x=1, in the above equation,}\Rightarrow y\times e^{ta{n}^{-1}x}=\frac{e^{(m+1)ta{n}^{-1}x}}{(m+1)}+C\Rightarrow 1\times e^{\frac{\pi }{4}}=\frac{e^{(m+1)\frac{\pi }{4}}}{(m+1)}+C\therefore C=\frac{e^{(m+1)\frac{\pi }{4}}}{(m+1)}-e^{\frac{\pi }{4}}$

Particular solution of the D.E. is $y\times e^{ta{n}^{-1}x}=\frac{e^{(m+1)ta{n}^{-1}x}}{(m+1)}+\frac{e^{(m+1)\frac{\pi }{4}}}{(m+1)}-e^{\frac{\pi }{4}}$