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:
Which is the required solution. 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 \[y\cdot IF=\int{(Q\cdot IF)\,dx}+C\] \[yx=\int{\frac{-\,1}{x(1+{{x}^{2}})}\cdot x\,dx+C}\] \[\Rightarrow \] \[yx=\int{\frac{-\,1}{1+{{x}^{2}}}dx+C}\] \[\Rightarrow \] \[yx={{\cot }^{-1}}x+C\] \[\left[ \because \,\,\int{\frac{-\,1}{1+{{x}^{2}}}dx={{\cot }^{-1}}x} \right]\] \[\therefore \] \[y=\frac{{{\cot }^{-1}}x+C}{x}\]
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