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muso07
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Homework Statement
Find the general solution of the following differential equation.
y'+4xy=10xy2cos(x2)
Homework Equations
The usual Bernoulli equation ones:
y'+p(x)y = g(x)ya
u(x)=[y(x)]1-a
u'+(1-a)pu = (1-a)g
The Attempt at a Solution
I got up until the general solution part.. I'll just type bits of it out because it'll take me ages (I'm a slow typer).
So in the equation, a=2, p(x)=4x, g(x)=10xcos(x2)
Change of variables:
u(x)=[y(x)]1-a = y-1
and u'+(1-a)pu = (1-a)g
=> u'-4xu = -10xcos(x2)
Now here's where I get confused..
General solution:
u= e[tex]\int[/tex]p(x)dx[[tex]\int[/tex]r(x)e[tex]\int[/tex]p(x)dxdx+c
where p=-4x, r=-10xcos(x2)
u= e[tex]\int[/tex]-2x^2[[tex]\int[/tex](-10xcos(x2)e[tex]\int[/tex]-2x^2)dx+c
Argh.. that's messy, I hope it makes sense.
Anyway, I can't seem to figure out that integral... I've used parts and stuff but I don't seem to be getting anywhere.
Any help would be appreciated.
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