Undetermined coefficients + initial value problem

[becca4]

Homework Statement

Find solution for y'' + y' -2y = 2x with initial values of y(0) = 0, y'(0) = 1

Homework Equations

I have found yc = c1*exp(-2x) + c2*exp(x), but finding yp is what I'm having trouble with... AND THEN I'm not so sure how to go about the initial value.

The Attempt at a Solution

Ok, so my "reasonable assumption" :
yp = Ax
y'p = A
y''p= 0

When I plug into equation, I get:
0 + A -2(Ax) = 2x... and that's where I'm stuck.

Any insights?

 

[EnumaElish]

Is yc the general solution and yp the particular?

If so, shouldn't you be using the initial conditions to solve for c1 and c2?

 

[becca4]

Edit: general solution y = yc + yp.

Does that help?

My strategy is to find the yc part of the general y solution, pretending like the left side of the original eqn is equal to zero. Then to find yp I'm supposed to make a reasonable assumption to find it. All parts of yc and yp must be linearly independent, so yp must be in a similar form of the right side of the original equation. So, I take yp, take 1st and 2nd derivatives and see if whatever I get agrees with 2x. Then when i find what yp is, I add it to yc.

...and that's all I know about this equation.

 

[Dick]

Your "reasonable assumption" is reasonable, but not general enough. Try yp=Ax+B.

 

[HallsofIvy]

In general, for a RHS of xⁿ, you will need y_p= Axⁿ+ Bx^n-1+ Cx^n-2 . . .

 

[becca4]

Your "reasonable assumption" is reasonable, but not general enough. Try yp=Ax+B.

Ahhh... Clever.

 

[becca4]

It worked! Thanks much!

Now with the initial values:
I have my general solution y = c1*exp(-2x) + c2*exp(x) -x -1/2

All I'll do to find c1 and c2 is use y(0)=0 in that equation, then take its derivative and use the y'(0)=1, that's the way to go, correct?

 

[becca4]

In general, for a RHS of xⁿ, you will need y_p= Axⁿ+ Bx^n-1+ Cx^n-2 . . .

Gotcha.