Initial value inhomogeneous differential equation

[arl146]

Homework Statement

In this problem, we will solve the initial value inhomogeneous differential equation in two steps.
y''+4y=2x*e^(3x) y(0)=0 y'(0)=0

1.) First determine y_p the particular solution to the differential equation.
2.) Next determine y=y_c+y_p the complete solution to the differential equation.

The Attempt at a Solution

the roots i got were 0 and -4.
i did y''+4y=0
'equivalent' to: r^2 + 4r = 0
dont know what to do next. usually we'll go onto the y_c part but idk what its supposed to look like. the notes suggest that equation will have sin(something) and a cos(something)

help please! i should be able to do part 2 on my own; but we'll see

 

[Mark44]

Homework Statement

In this problem, we will solve the initial value inhomogeneous differential equation in two steps.
y''+4y=2x*e^(3x) y(0)=0 y'(0)=0

1.) First determine y_p the particular solution to the differential equation.
2.) Next determine y=y_c+y_p the complete solution to the differential equation.

The Attempt at a Solution

the roots i got were 0 and -4.

No. The roots of the equation r² + 4 = 0 are not 0 and -4.

i did y''+4y=0
'equivalent' to: r^2 + 4r = 0
dont know what to do next. usually we'll go onto the y_c part but idk what its supposed to look like. the notes suggest that equation will have sin(something) and a cos(something)

help please! i should be able to do part 2 on my own; but we'll see

 

[arl146]

oh... i see what i did wrong there. i see it is 4 instead of 4r. the roots are +2i and -2i

 

[Mark44]

So the solution to the homogeneous problem will involve sin(2x) and cos(2x).

Do you have any thoughts on a particular solution?

 

[arl146]

what do you mean homogenous? the problem calls is an inhomogenous one?

but, an example in the book shows y''+4y= e^(3x). when they did yp, the did yp=Ae^(3x)
but how would mine be since i have that 2x multiplied by my e^(3x) ?

 

[Mark44]

The related homogeneous problem is y'' + 4y = 0. sin(2x) and cos(2x) are solutions to that problem.

If the nonhomogeneous problem happened to be y'' + 4y = 1, you would try a particular solution of y_p = A, and then you would figure out what A needed to be.

If the problem happened to be y'' + 4y = 2x, you would try a particular solution of y_p = A + Bx.

If it were y'' + 4y = 5x², you would try a particular solution of y_p = A + Bx + Cx².

The example in your book was y'' + 4y = e^3x, so you try y_p = Ae^3x.

Now with a right side of 2xe^3x, what might you try for a particular solution?

 

[arl146]

i don't know =[ ... Ae^(3x)+Bx?

 

[Mark44]

How about (A + Bx)e^(3x)?

 

[arl146]

Yea ok that makes sense. So what now.. I set A= something and Bx=something to get the value if A and B right?

 

[lurflurf]

You need to find a and b such that
y=(A + Bx)e^(3x) is a solution of the equation
that is
y''+4y=2x*e^(3x)
or
[(A + Bx)e^(3x)]''+4(A + Bx)e^(3x)=2x*e^(3x)

 

[arl146]

so it would be:
3e^(3x) * (2B+3A+3Bx) + 4*(A+Bx)e^(3x) = 2x*e^(3x)
after some simplifying i ended up getting 6B+13A=0 and 13Bx=2x
so B= 2/13 and A= -12/169

 

[Mark44]

so it would be:
3e^(3x) * (2B+3A+3Bx) + 4*(A+Bx)e^(3x) = 2x*e^(3x)
after some simplifying i ended up getting 6B+13A=0 and 13Bx=2x
so B= 2/13 and A= -12/169

You don't need to ask us - you can check it for yourself. Is (-12/169 + 2/13*x)e^(3x) a solution of y'' - 4y = 2xe(3x)?

 

[arl146]

i got the answer actually. thanks for all the help !