# Thread: Another Differential Equation Question

1. ## Another Differential Equation Question

Hi again, here's another question that I had a go at but haven't got any answers for, does the working and answer seem ok?

Solve $y' + y = y^2e^x$

The method I use is outlined in my lecture notes, however I am unsure of it's actual name.

First thing I did was to isolate the x term on the right hand side by multiplying everything by $y^{-2}$, giving the equation $y'y^{-2} + y^{-1} = e^x$

I then use the substitution of $w = y^{-1}$, then differentiate this equation with respect to x and you get $w' = -y^{-2}y'$.

Placing these two substitutions into my original equation this simplifies to be $w' - w = -e^x$.

I wasn't too sure what to do at this point, I took a guess at using an integrating factor of $e^{\int{-1}}$, which when I put back into the equation gave me $e^{-x}w' - we^{-x} = -1$.

This then integrates to $we^{-x} = x$, which after putting back in my substitution gave the answer of $xye^{x} = 1$.

Is any of this correct, or know me have I made an early error and then just answered a completely different question?

Thanks again

Craig

2. Originally Posted by craig
Hi again, here's another question that I had a go at but haven't got any answers for, does the working and answer seem ok?

Solve $y' + y = y^2e^x$

The method I use is outlined in my lecture notes, however I am unsure of it's actual name.

First thing I did was to isolate the x term on the right hand side by multiplying everything by $y^{-2}$, giving the equation $y'y^{-2} + y^{-1} = e^x$

I then use the substitution of $w = y^{-1}$, then differentiate this equation with respect to x and you get $w' = -y^{-2}y'$.

Placing these two substitutions into my original equation this simplifies to be $w' - w = -e^x$.

I wasn't too sure what to do at this point, I took a guess at using an integrating factor of $e^{\int{-1}}$, which when I put back into the equation gave me $e^{-x}w' - we^{-x} = -1$.

This then integrates to $we^{-x} = x$, which after putting back in my substitution gave the answer of $xye^{x} = 1$.

Is any of this correct, or know me have I made an early error and then just answered a completely different question?

Thanks again

Craig
I think you'll find that $w = -x e^{x} + C e^x = e^x (C - x)$.

Then $y = \frac{e^{-x}}{C - x}$.