question_answer

Solve the following initial value problem.

\[2xy+{{y}^{2}}-2{{x}^{2}}\frac{dy}{dx}=0,\] \[y(1)=2\]

OR

Solve the following differential equation.

\[(1+y+{{x}^{2}}y)\,dx+(x+{{x}^{3}})dy=0\]

Answer:

We have,

\[2xy+{{y}^{2}}-2{{x}^{2}}\frac{dy}{dx}=0\] \[\Rightarrow \] \[2{{x}^{2}}\frac{dy}{dx}=2xy+{{y}^{2}}\]

\[\Rightarrow \]   \[\frac{dy}{dx}=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}\]

Which is a homogeneous differential equation.

Putting y = vx and \[\frac{dy}{dx}=v+x\frac{dv}{dx},\] it reduces to

\[v+x\frac{dv}{dx}=\frac{2v+{{v}^{2}}}{2}\]

\[\Rightarrow \]   \[2v+2x\frac{dv}{dx}=2v+{{v}^{2}}\]

\[\Rightarrow \]   \[2x\frac{dv}{dx}={{v}^{2}}\]

\[\Rightarrow \]   \[\frac{2dv}{{{v}^{2}}}=\frac{dx}{x}\]

\[\Rightarrow \]   \[2\int{\frac{1}{{{v}^{2}}}dv}=\int{\frac{1}{x}dx}\]

\[\Rightarrow \]   \[\frac{-\,2}{v}=\log \,\,x+C\]

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

It is given that y(1) = 2, i .e. y = 2 when x = 1.

Putting x = 1 and y = 2 in Eq. (i), we get

\[-\,1=0+C\,\,\,\,\Rightarrow \,\,\,\,C=-\,1\]

Putting \[C=-\,1\] in Eq. (i), we get

\[\frac{-\,2x}{y}=\log |x|-\,1\]

\[\Rightarrow \]   \[y=\frac{2x}{1-\log \,\,x}\]

Hence, \[y=\frac{2x}{1-\log \,\,x}\] gives the solution of given differential equation.

OR

Given differential equation is

\[(1+y+{{x}^{2}}y)dx+(x+{{x}^{3}})dy=0\]

\[\Rightarrow \]   \[(x+{{x}^{3}})dy=-\,(1+y+{{x}^{2}}y)dx\]

\[\Rightarrow \]   \[\frac{dy}{dx}=\frac{-\,[1+y(1+{{x}^{2}})]}{x+{{x}^{3}}}\]

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

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

\[\Rightarrow \]                 \[\frac{dy}{dx}+\frac{y}{x}=\frac{-\,1}{x(1+{{x}^{2}})}\]                   ?(i)

Which is a linear differential equation of the form

\[\frac{dy}{dx}+Py=Q\]                                      ?(ii)

where, \[P=\frac{1}{x}\] and \[Q=\frac{-\,1}{x(1+{{x}^{2}})}\]

Now,     \[IF={{e}^{\int{Pdx}}}={{e}^{\int{\frac{1}{x}dx}}}={{e}^{\log \,\,x}}=x\]

Hence, its solution is given by

\[y\cdot IF=\int{(Q\cdot IF)\,dx}+C\]

\[yx=\int{\frac{-\,1}{x(1+{{x}^{2}})}\cdot x\,dx+C}\]

\[\Rightarrow \]   \[yx=\int{\frac{-\,1}{1+{{x}^{2}}}dx+C}\]

\[\Rightarrow \]   \[yx={{\cot }^{-1}}x+C\]                                           \[\left[ \because \,\,\int{\frac{-\,1}{1+{{x}^{2}}}dx={{\cot }^{-1}}x} \right]\]

\[\therefore \]      \[y=\frac{{{\cot }^{-1}}x+C}{x}\]

Which is the required solution.