Answer to Question #102571 in Differential Equations for Ajay

Question

Apply the method of variations of parameters to solve the following differential equations:x^2y" +xy'-y=x^2e^x

Answer 1

y''+p(x)y'+q(x)y=r(x)

$yp=-y1\int(y2*r(x))/(W(y1,y2))dx+y2\int(y1*r(x))/(W(y1,y2))dx$

y''+y'/x-y/x^2=exp[x]

y1=x

y2=1/x

$W(y1,y2)=-2/x$

$yp=-x\int((1/x)*exp[x]/(-2/x))dx+1/x\int((x*exp[x])/(-2/x))dx$

$yp=x\int(exp[x]/2)dx-1/(2x)\int((x^2*exp[x])dx$

$yp=x/2(exp[x]+c1)-1/(2x)*(exp[x]*(2-2x+x^2)+c2)$

$y=exp[x]-exp[x]/x+c1*x+c2/x$