question_answer

Find the particular solution of the following differential equation : \[xy\frac{dy}{x}=(x+2)(y+2);y=-\,1,\] when x = 1. Or Find the particular solution of the following differential equation.   \[x({{x}^{2}}-1)\frac{dy}{dx}=1;\] y = 0, when x = 2

Answer:

We have,

\[xy\frac{dy}{dx}=(x+2)\,(y+2)\Rightarrow \frac{y}{y+2}dy=\frac{x+2}{x}\,dx\]

\[\Rightarrow \]   \[\left( 1-\frac{2}{y+2} \right)\,dy=\left( 1+\frac{2}{x} \right)\,\,dx\]

On integrating, we get

\[y-2\,\,\log \,\,(y+2)=x+2\,\,\log x+C\]?(i)

Given,\[y=-\,1,\] when \[x=1,\]then from Eq. (i),

\[-\,1-2\,\log \,(-\,1+2)=1+2\,\log 1+C\]

\[[\because \log 1=0]\]

\[\Rightarrow \]   \[C=-\,2,\]

Then eq. (i) becomes

\[y-2\,\,\log \,\,(y+2)=x+2\,\,\log \,\,x-2\]

\[\Rightarrow \]   \[y=x-2+2\,\{\log \,(y+2)+\log x\}\]

\[\Rightarrow \]   \[y=x-2+2\,\,\log \,\,\{x\,(y+2)\}\]

Which is the required particular solution.

OR

We have, \[x\,({{x}^{2}}-1)\frac{dy}{dx}=1\Rightarrow dy=\frac{1}{x\,({{x}^{2}}-1)}\,dx\]

Put \[({{x}^{2}}-1)=t\Rightarrow 2xdx=dt\]

Now,\[\int{dy=\frac{1}{2}}\int{\frac{dt}{t\,(t+1)}}\]

Resolving into partial fractions, we get

\[\int{dy=\frac{1}{2}}\left[ \int{\,\left( \frac{1}{t}-\frac{1}{t+1} \right)\,dt} \right]\]

\[\Rightarrow \]   \[y=\frac{1}{2}[\log t-\log |t+1|]+C\]

\[\Rightarrow \]   \[y=\frac{1}{2}[\log t-\log [t+1]]+C\]

\[\Rightarrow \]   \[y=\frac{1}{2}\log \left| \frac{t}{t+1} \right|+C\]                         ?(i)

\[\Rightarrow \]   \[y=\frac{1}{2}\log \left| \frac{{{x}^{2}}-1}{{{x}^{2}}} \right|\]

When \[x=2\]then\[y=0\]

\[\therefore \]      \[C=-\frac{1}{2}\log \frac{3}{4}\]

From Eq. (i), \[y=\frac{1}{2}\log \,\left( 1-\frac{1}{{{x}^{2}}} \right)-\frac{1}{2}\log \frac{3}{4}\]

which is the required particular solution.