find the general solution equation for:
y'+2y=2t(e^2t)
with initial condition: y(0)=1

Question

find the general solution equation for:
y'+2y=2t(e^2t) 
with initial condition:  y(0)=1

amistre64 · Answer

you have a characteristic equation of the homogenous part as:

r+2=0;  therefore r=-2$$y_h=c~e^{-2t}$$

anonymous · Answer

well my professor said to use integration factor

amistre64 · Answer

or you can use e^(2t) as an integrating thingy

anonymous · Answer

haha ya i used that and got $$e ^{2y}t=\int\limits e ^{2t} (2te ^{2t})dt$$

anonymous · Answer

left side is supposed to be with y not t **

amistre64 · Answer

$$\int e^{2t}(y'+2y=2t~e^{2t})~dt $$
$$ e^{2t}y=2\int t~e^{4t}~dt $$

anonymous · Answer

ya i got that but im not sure if i have to use integration by parts or not

amistre64 · Answer

yes, you need to.  unless you have remember this setup from the tables ....

amistre64 · Answer

$$\int x~e^{ax}dx=e^{ax}(\frac{1}{a}x-\frac{1}{a^2})$$
should be right

anonymous · Answer

ok so it would be $$=\frac{t ^{4} }{ 2 }e ^{2t}-e ^{2t}/4$$

amistre64 · Answer

umm, no

amistre64 · Answer

close tho

amistre64 · Answer

$$e^{2t}y=e^{4t}(\frac t{4}-\frac{1}{16})+C$$

anonymous · Answer

o lol i see where i messed up

amistre64 · Answer

:)  then of course its left with algebra, divide off the e^{2t}  and insert your initial condition to determine a suitable C

anonymous · Answer

yep got it

amistre64 · Answer

good job ... you find out that its not so much the calculus that gets you; its all the little mistakes we tend to do in the algebra that kills us lol