Question

y''+4y=t^(2)+3e^t

find the general solution of the nonhomogenous differential equation

Answer 1

find the homogenous part

then add to it a guessed at nonhomogenous (particular) part

Answer 2

what is our characteristic equation to root out?

Answer 3

its gonna be complex im sure ....

Answer 4

r^2 + 4 = 0

r = 0 +- 2i

right?

Answer 5

\[y_h=c_1cos(2x)+c_2sin(2x)\]
then we have to form a nonhomogenous from the form given above

Answer 6

and x=t ...

Answer 7

given: t^(2)+3e^t

\[y_p=(Ax^2+B)+(Ce^t)\]

since none of these are a copy of what we found before, there doesnt need to be any adjustments

Answer 8

\[y_p=Ax^2+B+Ce^t\]
\[(y_p)'=2Ax+Ce^t\]
\[(y_p)''=2A+Ce^t\]

\[(y_p)''+4(y_p)=2A+Ce^t+4(Ax^2+B+Ce^t)=(1)t^2+(3)e^t\]

simplify and compare coeefs

Answer 9

\[y_p=Ax^2+Bx+C+De^t\]actually

Answer 10

helps if i keep my ts in place instead of introducing xs tho ...

\[2A+Ce^t+4Ax^2+4B+4Ce^t=(1)t^2+(3)e^t\]
\[(4A)t^2=1t^2\]
\[5Ce^t=(3)e^t\]
\[(2A+4B)=0 \]

we got this in order?

Answer 11

A = 1/4
C = 3/5

2(1/4) + 4B = 0
4B = -1/2
B = -1/8

Answer 12

... so much typing and ugh ... yeah, x^2+x+k

Answer 13

i still have to do these on paper lol

Answer 14

you got the right answer though
haha we both have x's and t's all mixed together

Answer 15

once its done correctly :)

\[Y=y_h+y_p\]

Answer 16

thanks :)