Question

power series soln of xy'-(x+2)y-2x^2-2x=0

Answer 1

Divide original equation through by x and rearrange:

y'-(x+2)y/x=2x+2, y'-(1+2/x)y=2x+2.

Find integrating factor: e^{∫-(1+2/x)dx}=e^-x-2ln(x)=e^-xe^-2ln(x)=e^-x/x². (-2ln(x)=-ln(x²)=ln(1/x²); e^ln(1/x²)=1/x²)

Multiply through by e^-x/x²:

e^-xy'/x²-(x+2)ye^-x/x³=2e^-x/x+2e^-x/x².

[Note that the derivative of e^-x/x² wrt x is found using the product rule:

e^-x/x²=e^-xx^-2; let u=e^-x, so du/dx=-e^-x; v=x^-2, so dv/dx=-2x^-3=-2/x³;

d(uv)=udv/dx+vdu/dx=-2e^-x/x³-e^-x/x²=-(e^-x/x²)(2/x+1)=-(e^-x/x²)(x+2)/x=-(x+2)e^-x/x³.]

e^-xy'/x²-(x+2)ye^-x/x³=d(e^-xy/x²)/dx=2e^-x/x+2e^-x/x²,

e^-xy/x²=2∫(e^-x/x+e^-x/x²)dx=2(I+J)=

where I=∫(e^-x/x)dx and J=∫(e^-x/x²)dx.

Let u=1/x, then du/dx=-1/x²; dv=e^-xdx, then v=-e^-x;

I=-e^-x/x-∫(e^-x/x²)dx=-e^-x-J; 

So I+J=-e^-x/x+C₁ (C₁ is an arbitrary constant of integration),

e^-xy/x²=-2e^-x/x+2C₁, y/x²=-2/x+Ce^x, y=-2x+Cx²e^x, y'=-2+2Cxe^x+Cx²e^x, xy'=-2x+2Cx²e^x+Cx³e^x; (C=2C₁)

-(x+2)y=-xy-2y=2x²-Cx³e^x+4x-2Cx²e^x;

xy'-(x+2)y-2x²-2x=-2x+2Cx²e^x+Cx³e^x+2x²-Cx³e^x+4x-2Cx²e^x-2x²-2x=0.

The terms with the same colour cancel to leave no terms.

Therefore y=Cx²e^x-2x is the general solution where C is an arbitrary constant.

e^x=1+x+x²/2!+x³/3!+...+xⁿ/n! and Cx²e^x=C∑xⁿ⁺²/n!-2x, where 0≤n≤∞.