Solve general differential equation: please check my work?

[darryw]

Homework Statement

I forgot how to do these and just need a little refreshing on how to solve these

Homework Equations

y' - 2y = (t^2)(e^2t)

The Attempt at a Solution

y' - 2y = (t^2)(e^2t)

ok so my integrating factor is the value in front of y, right?
so mu(x) = e^int-2
= e^-2t

so then after i multiply whole equation by e^-2t i get:
(e^-2t)y = t^2 (because the e^-2t canceled the e^2t that was in the right hand term).


SO this is where I totally forgot how to proceed! Do i integrate both sides, then I am done? or am i missing a step?

and if i do integrate then do i use u-sub for the term on LHS?
thanks for any help

 

[rock.freak667]

Remember your LHS always becoms

d/dt( y*integratingn factor)


so now you just integrate both sides.

 

[darryw]

integ (ye^-2t)' = integ t^2

= ye^-2t = (1/3)t^3

y = (t^3) / (3e^-2t) + c (this is my final solution)

Is this correct? Also, even though I don't have initial cond can i say anything more about what c is?

thanks

 

[rock.freak667]

integ (ye^-2t)' = integ t^2

= ye^-2t = (1/3)t^3

y = (t^3) / (3e^-2t) + c (this is my final solution)

Is this correct? Also, even though I don't have initial cond can i say anything more about what c is?

thanks

The line in red should be

ye^-2t+(1/3)t³+C

then you can divide by e^-2t

 

[darryw]

thanks for the reply..but i don't understand what and why you did that? what does that equal? If zero, then how did (1/3)t^3 become negative?

 

[rock.freak667]

thanks for the reply..but i don't understand what and why you did that? what does that equal? If zero, then how did (1/3)t^3 become negative?

Terribly sorry, I made a typo, it should be this.

ye^-2t=(1/3)t³+C

 

[darryw]

ok thanks. so only diff was that i need to reference c right after integration before doing anything else to equation, --right?

 

[Mark44]

integ (ye^-2t)' = integ t^2

= ye^-2t = (1/3)t^3

In the line above, the first = shouldn't be there, and you should add the constant of integration in this step.

Each line should be an equation that implies, but is not equal to, the next line. After all, an equation can't be equal to another equation.

To recap,
d/dt(ye^{-2t} = t^2
\Rightarrow ye^{-2t} = \int t^2 dt = \frac{t^3}{3} + C
\Rightarrow y = e^{2t}(\frac{t^3}{3} + C)}

Notice also that the constant is multiplied by e^2t.

y = (t^3) / (3e^-2t) + c (this is my final solution)

Is this correct? Also, even though I don't have initial cond can i say anything more about what c is?

thanks

 

[darryw]

thanks for the replies..
so my final solution is:

y = (1/3) (t^3)((e^t)^2) + ce^t^2

is this correct? also, since there were no init cond, is this what's called the general solution to the DE? If i did have IC i would be able to know what c is right?
thanks

 

[Mark44]

thanks for the replies..
so my final solution is:

y = (1/3) (t^3)((e^t)^2) + ce^t^2

is this correct? also, since there were no init cond, is this what's called the general solution to the DE? If i did have IC i would be able to know what c is right?
thanks

No, not correct. See post #8. Where are you getting e^t2? And why are you writing e^2t as (e^t)²?

Yes, if you had an initial condition you could solve for C.

 

[darryw]

should be:

y = (1/3)(e^2t)(t^3) + c(e^2t) (i integrated integrating factor incorrectly)

is this correct? (also since there were no init cond, is this what's called the general solution to the DE? If i did have IC i would be able to know what c is right?
thanks)

 

[Mark44]

should be:

y = (1/3)(e^2t)(t^3) + c(e^2t) (i integrated integrating factor incorrectly)

is this correct?

Why are you asking this? Look at post #8.

(also since there were no init cond, is this what's called the general solution to the DE? If i did have IC i would be able to know what c is right?

Yes, this is the general solution. The other question has already been answered.