# Separable differential equations problems

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• Aug 13th 2007, 03:36 AM
Obstacle1
Separable differential equations problems
Two problems I'm stuck on.( sorry these aren't separable, they are first order linear, i wrote the wrong thing)

1: t*(dy/dt) + 2y = t^2 - t + 1

2: t*(dy/dt) + (t+1)*y = t

Any help finding the general solutions would be much appreciated.
I already got the integrating factors as 1: exp(2lnt) and 2: exp(t+lnt)
• Aug 13th 2007, 05:08 AM
galactus
For #1:

$\displaystyle ty'+2y=t^{2}-t+1$

Divide by t: $\displaystyle y'+\frac{2}{t}y=t-1+\frac{1}{t}$

The integrating factor is 2/t.

$\displaystyle e^{\int{\frac{2}{t}}dt}=t^{2}$

$\displaystyle \frac{d}{dt}[t^{2}y]=t^{2}(t-1+\frac{1}{t})=t^{3}-t^{2}+t$

Integrate:

$\displaystyle t^{2}y=\frac{t^{4}}{4}-\frac{t^{3}}{3}+\frac{t^{2}}{2}+C$

Solve for y:

$\displaystyle y=\frac{t^{2}}{4}-\frac{t}{3}+\frac{1}{2}+Ct^{-2}$
• Aug 13th 2007, 05:42 AM
Krizalid
Quote:

Originally Posted by Obstacle1
Two problems I'm stuck on.( sorry these aren't separable, they are first order linear, i wrote the wrong thing)

1: t*(dy/dt) + 2y = t^2 - t + 1

2: t*(dy/dt) + (t+1)*y = t

Any help finding the general solutions would be much appreciated.
I already got the integrating factors as 1: exp(2lnt) and 2: exp(t+lnt)

When you have a linear ODE like $\displaystyle y'+P(x)y=Q(x)$ the integrating factor is given by $\displaystyle \color{blue}\mu(x)=\exp\int P(x)\,dx$

Then you amplify the equation by $\displaystyle \mu$ to set the left side into a product's function derivative, finally all you have to do is integrate.
• Aug 13th 2007, 10:50 AM
Krizalid
Quote:

Originally Posted by Obstacle1
I already got the integrating factors as 2: exp(t+lnt)

$\displaystyle \exp(t+\ln t)=\exp(\ln e^t+\ln t)=t\cdot e^t$

That's all.