Ordinary Differential Equations/Exact 4

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Q1 answer:dydx+2y=x2e2x+5y=eP(x)dxQ(x)dx+CeP(x)dxeP(x)dxP(x)=2e2dx=e2xQ(x)=x2e2x+5y=(e2x)(x2e2x+5)dx+Ce2xy=x2+5e2xdx+Ce2x=x33+5e2x2+Ce2xy=x33e2x+52+Ce2x_ Template:BookCat