Find the solution to the differential equation y''-1y'-6y=-30x^2+8x+19 initial conditions y(0)=0 and y'(0)=-15

Question

trying to do homework but cant figure out this question. cant anyone help please?

Answer 1

This is a nonhomogeneous equation. To find the general solution, we should first find the solution of the corresponding homogeneous equation, and then shift it by one particular solution of the nonhomogeneous equation.

So first, we solve y'' - 1y' - 6y = 0. To do this, we factor the characteristic polynomial, a^2 - a - 6 = (a - 3)(a + 2). So the most general solution of this equation is Ae^(-2x) + Be^(3x).

Now we have to find one particular solution to the nonhomegenous version, y'' - 1y' - 6y = -30x^2 + 8x + 19. Let's assume that y = Px^2 + Qx + R and see what we come up with:

y'' - 1y' - 6y = -30x^2 + 8x + 19

2P - 1(2Px + Q) - 6(Px^2 + QX + R) = -30x^2 + 8x + 19

2P - 2Px - Q - 6Px^2 -6Qx -6R = -30x^2 +8x + 19

-6Px^2 + (-2P -6Q)x + (2P - Q - 6R) = -30x^2 + 8x + 19

Since we know the coefficients have to match up, then now we can solve for P, Q, R:

-6P = -30
-2P - 6Q = 8
2P - Q - 6R = 19

From the first equation, P = 5. Let's sub that into the next two:

-10 - 6Q = 8
10 - Q - 6R = 19

Now we know Q = -3. Let's sub that into the last equation:

10 + 3 - 6R = 19

Therefore, R = -1.

This means that one particular solution to y'' - 1y' - 6y = -30x^2 + 8x + 19 is the polynomial 5x^2 - 3x - 1. Therefore, the general solution is:

y = Ae^(-2x) + Be(3x) + 5x^2 - 3x - 1.

Now we're ready to plug in the initial conditions. We have y(0) = 0. Plugging this in gives

0 = A + B - 1.

We also know y'(0) = -15. Well, differentiating y gives

y' = -2Ae^(-2x) + 3Be^(3x) + 10x - 3.

Since y'(0) = -15, we have

-15 = -2A + 3B - 3.

We now have the following system:

0 = A + B - 1
-15 = -2A + 3B - 3

Let's solve via substitution. The first equation tells us that A = 1 - B. Subbing this in to the second equation gives -15 = -2(1 - B) + 3B - 3, or -15 = -2 + 2B + 3B - 3, or -10 = 5B, so B = -2. Then A = 3. So the final solution is

y = 3e^(-2x) - 2e^(3x) + 5x^2 - 3x - 1.

Answer 2

The characteristic equation is r^2 - r - 6 = 0, which gives (r-3)(r+2) = 0, so r = 3 or -2. So you have Ae^(3x) + Be^(-2x).

Now the next step can probably be done in various ways.. I'd just let y = Cx^2 + Dx + E and solve for C,D,E:
y' = 2Cx + D
y'' = 2C
2C - (2Cx + D) - 6(Cx^2 + Dx + E) = -30x^2 + 8x + 19

So 2C-D-6E = 19, -2C-6D = 8, -6C = -30.
So C = 5, D = -3, E = -1.

So the general solution is y = 5x^2 - 3x - 1 + Ae^(3x) + Be^(-2x).

We also have y' = 10x - 3 + 3Ae^(3x) - 2Be^(-2x).

Substituting in the initial conditions:
0 = -1 + A + B
-15 = -3 + 3A - 2B.
Solving gives A = -2, B = 3.

So the solution is y = 5x^2 - 3x - 1 - 2e^(3x) + 3e^(-2x)

Answer 3

0.. i dunno i aint chinese...lol j/k