Answer to Question #204122 in Differential Equations for rikajk

Question #204122

HOMOGENEOUS EQUATION

Solve the given equation below. Show complete solution.


xy' = x + y


1
Expert's answer
2021-06-07T18:43:13-0400

Ans:-

xy=x+yxy'=x+y


dydx=x+yx           ...1\dfrac{dy}{dx}=\dfrac{x+y}{x}\ \ \ \ \ \ \ \ \ \ \ ...1


Solving dydx\dfrac{dy}{dx} by putting   y=vx\ \ y=vx


differentiating w. r. t. x


dydx=xdvdx+vdxdx\dfrac{dy}{dx}=x\dfrac{dv}{dx}+v\dfrac{dx}{dx}


dydx=xdvdx+v\dfrac{dy}{dx}=x\dfrac{dv}{dx}+v


Putting value of dydx\dfrac{dy}{dx} and y=vx in (1)


dydx=x+yx\dfrac{dy}{dx}=\dfrac{x+y}{x}


xdvdx+v=x+vxxx\dfrac{dv}{dx}+v=\dfrac{x+vx}{x}\\


xdvdx+v=1+vx\dfrac{dv}{dx}+v=1+v\\


dvdx=1x\dfrac{dv}{dx}=\dfrac{1}{x}


Integrating both sides


dv=dxx\int dv=\int \dfrac{dx}{x}


v=logx+cv=log|x| +c


putting v=yxv=\dfrac{y}{x}


yx=logx+c\dfrac{y}{x}=log|x| +c


y=xlogx+cxy=xlog|x|+cx


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