How do I solve the following non-homogenous differential equation?

Question

d^2y/dx^2 +2dy/dx - 8y =4

Thank you for any help.

Answer 1

convert it to
D^2+2D -8 =4

solve for the complementary sol'n

m^2+2m-8=0
(m-2)(m+4)=0; m=-4, 2
yc=c1*e^(-4t)+c2*e^(2t)

solve for the particular sol'n by method of undetermined coefficients
let yp=K
take the first and second derivatives of yp
D''(yp)=D'(yp)=0


subsitute this to the equation
(D^2+2D -8)yp =4
D^2(yp)+2D(yp)-8(yp)=4
0 + 0 -8yp=4
yp=-2

then y= yp+yc =c1*e^(-4t)+c2*e^(2t)-2

Answer 2

I just relearned how to do these...so yeah

Anyways, i forget hte exact terms, but first find the homogenous solution

y''+2y' - 8 = 0
r^2+ 2r - 8 =0
(r+4)(r-2)
so your roots are r=-4 and r = 2
You get
y= c1e^(-4t) + c2e^(2t)
Now just put this equation aside

Hmm..how to word this

Solve for the polynomial on the right side of the equation.
in this case its t^0 * 4 = 4
We're expecting a soultion that looks similar to this.....

I'll give a second example at the bottom to help you see what i'm talking about

but in thise case....we want a constant. So
let y = A......A is a constant
find y' and y''.....so take the derivative of y=A with respect to t
y' = 0
y'' = 0
Now plug them into your second order equation

0+ 2*0 - 8 *A = 4
so thats
-8*A = 4
A = -1/2
So y= -1/2

now that equation i told you to forget about above
..just mesh these two equations together

y= c1e^(-4t) + c2e^(2t) - 1/2

here is another example...if it ends up being too confusing..then just ignore it.
************************
ok pretend instead we have y''+2y' -8y = t^2
the first part we did above is the same
the second part now is.......remember we're trying to find a similar solution
y=A*t^2 + Bt +C...... then y' = 2At+B and y'' = 2A
then we'd do the same thing we did above..plug into the differential equation and then mesh the 2 equations together.

2A + 4A*t + 2B - 8 *At^2 -8Bt -8C = t^2

Realize t^2 = t^2 + 0*t+0

we combine the same ....umm..powers
So all the t^2 on the left side are equal to the t^2 on the right side
the t on the left = t on the right
constants on the left are equal to the constants on the right

-8*At^2 = t^2

4A*t - 8B*t = 0*t

2A +2B-8C = 0
so
-8A*t^2 = t^2
A= -1/8

4A*t - 8B*t = 0*t = 0
4A*t = 8B*t

-1/2 = 8B
B = -1/16

2A +2B-8C = 0
-2/8 -2/16 = 8C
-6/16 = 8C
C= - 3/64
so y = -1/8t^2 - 1/16 * t - 3/64...then the mesh

y= c1e^(-4t) + c2e^(2t) -1/8t^2 - 1/16 * t - 3/64..

Answer 3

The general solution of the corresponding homogeneous equation is

y = Ae^(2x) + Be^(-4x).

By the method of variation of parameters, we find a particular solution of the non-homogeneous equation by assuming a solution of the type

z = C(x)*e^(2x) + D(x)*e^(-4x) ,

where C(x) and D(x) are functions to be determined. Proceeding in the usual way, we obtain the following system of equations involving the derivatives of C and D.

C'*e^(2x) + D'*e^(-4x) = 0
2C'*e^(2x) - 4D"*e^(-4x) = 4 .

We find C' = (2*e^(-2x))/3 and D' = (-2*e^(4x))/12 ; therefore,
C = (-e^(-2x))/3 and D = (-2*e^(4x))/12 . We find z = -1/3-1/6
or z = -1/2.

The general solution is y + z = Ae^(2x) + Be^(-4x) - 1/2 .

Answer 4

x '' + 3x ' - 10x = t e^(t) The homogeneous eqn is r^2 + 3r - 10 = 0 => (r + 5)(r - 2) = 0 r = 2 and -5 the final answer is x(0) = A e^(2t) + Be^(-5t) enable particular answer is xp = (Ct + D)e^t x ' = Ce^t + (Ct + D)e^t = e^t (Ct + C + D) x '' = e^t (Ct + C + D) + Ce^t = e^t (Ct + 2C + D) substituting in the given DE e^t (Ct + 2C + D) + 3e^t(Ct + C + D) - 10e^t(Ct + D) = te^t e^t [Ct + 2C + D + 3Ct + 3C + 3-d - 10Ct - 10D ] = t e^t e^t [ -6Ct + 5C - 6D ] = t e^t te^t(-6C) + 5C - 6D = te^t -6C = a million ==> C = -a million/6 5C - 6D = 0 -6D = 5/6 D = -5/36 xp =[ (-a million/6)t - 5/36) ]e^t = - (a million/6)e^t - (5/36)e^t x(t) = x(0) + xp x(t) = A e^(2t) + Be^(-5t) - (a million/6)e^t - (5/36)e^t