Hi

I have attached my question

For part a) i got final answer as y=-4/5 +ce^(5/3x)

i got this using intergrating factors

However i cannot seem to do part b)

could someone please help

thank you

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- May 29th 2009, 06:12 AMmaster_2m8differential equation problem
Hi

I have attached my question

For part a) i got final answer as y=-4/5 +ce^(5/3x)

i got this using intergrating factors

However i cannot seem to do part b)

could someone please help

thank you - May 29th 2009, 06:42 AMartvandalay11
$\displaystyle x*y'+(1-2x)y=4x$ so $\displaystyle y'+\frac{1-2x}{x}y=4$

The integrating factor is then $\displaystyle e^{\int\frac{1-2x}{x}dx}$

So now, $\displaystyle \int\frac{1-2x}{x}dx=\int\left(\frac{1}{x}-\frac{2x}{x}\right)dx=ln|x|-2x$ and we ignore the +c...

So the integrating factor is $\displaystyle e^{ln|x|-2x}=e^{ln|x|}e^{-2x}=xe^{-2x}$

So $\displaystyle xe^{-2x}y'+e^{-2x}(1-2x)y=4xe^{-2x}$

and the left side of the equation is just $\displaystyle \frac{dy}{dx}(xe^{-2x}y)$

So $\displaystyle \frac{dy}{dx}(xe^{-2x}y)=4xe^{-2x}$

Now integrate both sides, but this time be sure to include the +c...

$\displaystyle xe^{-2x}y=-(2x+1)e^{-2x}+c$

and now you just have to use your initial condition to find c (that integration was done by parts which I trust you can do) - May 29th 2009, 06:50 AMCalculus26
See attachment