# Thread: Differential Equations... find the general sum

1. ## Differential Equations... find the general sum

dy/dt + 3y = t + e^(-2t)

Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf

So we did this problem in class and I'm a little confused about how the TA integrated it.

(e^(3t)y)' = te^(3t) + e^t
e^(3t)y = INT(te^(3t) + e^t)dt

u = t
du = dt
dv = e^(3t)dt
v = (1/3)e^(3t)

e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)

If someone could please explain how they got that after integration. Thank you.

2. Originally Posted by pakman
dy/dt + 3y = t + e^(-2t)

Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf

So we did this problem in class and I'm a little confused about how the TA integrated it.

(e^(3t)y)' = te^(3t) + e^t
e^(3t)y = INT(te^(3t) + e^t)dt

u = t
du = dt
dv = e^(3t)dt
v = (1/3)e^(3t)

e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)

If someone could please explain how they got that after integration. Thank you.

which are you confused about? the integration or the y = ... or both.
The TA did integration by parts to find the integral, and to go from
e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
to
y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)
he divided both sides by e^(3t)

which parts exactly do you want explained? do you know the method for integration by parts?

3. Originally Posted by pakman
dy/dt + 3y = t + e^(-2t)

Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf

So we did this problem in class and I'm a little confused about how the TA integrated it.

(e^(3t)y)' = te^(3t) + e^t
e^(3t)y = INT(te^(3t) + e^t)dt

u = t
du = dt
dv = e^(3t)dt
v = (1/3)e^(3t)

e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)

If someone could please explain how they got that after integration. Thank you.
It is of the form,

y'+P(t)y=Q(t) where P,Q are continous almost everywhere.

The integrating factor,

R(t)=exp (INT P(t) dt)=exp (3t) = e^{3t}

Thus, the solution is,

y= 1/R(t) *INT [R(t)*Q(t)] dt

=e^{-3t} * INT [e^{3t}*(e^{-2t}+t)] dt

=e^{-3t}*INT [t*e^{3t}+e^t ]dt

Now find the integral,

INT [t*e^{3t}+e^t] = INT t*e^{3t} dt +INT e^t dt

Integration by parts,

(1/3)te^{3t}-(1/9)e^{3t}+e^t+C

Thus,

y=e^{-3t} [ (1/3)te^{3t}-(1/9)e^{3t}+e^t+C]

=(1/3)t-(1/9)+e^{-2t}+C=(1/3)t+e^{-2t}+C

Or something like that, I am too lazy to check.