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I'm stuck with two problems that are aimed at introducing shortly the method of variation of parameters in order to solve a differential equation. The problems are:

x\cdot y(x)'+y(x)=x^2;\ y(1)=1

and

u'(t)+\frac{u}{1+t}=exp(2t);\ u(0)=4.

I have tried to understand the method, but I have not arrived anywhere yet. Can someone please help me get started?

-Marie :)

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    At what part of the method in particular are you stuck?2012-03-14

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For the first one, note that the left-hand side can be "factored" as (xy)', as it is just the result of the basic product rule. Can you integrate with a $+C$ and figure it out from there? For the second one, in keeping with the method of variation of parameters: what would you need to multiply both sides of the equation by so that the left-hand factors as a derivative of a product again? Symbolically, that means that q(t)u'+\frac{q(t)}{1+t}u (after multiplying by $q$) factors to \big(p(t)u\big)' for some $p$ and $q$. Expand out the latter with the chain rule and see if you can progress from there.

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    @MarieP: Good job! You forgot about the big $C$: plug in $t=0$ to find that $u(0)(1+0)=\frac{1}{4}e^{2(0)}(2\cdot0+1)+C \implies C=15/4.$ Put that into the solution and you have $u(t)=\frac{(2t+1)e^{2t}+15}{4}.$2012-03-15