Differential equation?

Question

y'' - 4y = (8x -2)(e^(2x))

How do you solve it?

Answer 1

1. Find the homogeneous solution, by setting the right hand side to zero.
yh'' - 4yh = 0
which is a linear DE with constant coefficients. The solution has the form:
yh = c × exp{λx}
Insert this in the DE divide by c × exp{λx} and get the characteristic polynomial:
λ² - 4 = 0
with the roots:
λ1 = -2 and λ2 = 2
Hence:
yh = c1 × exp{-2x} + c2 × exp{2x}

2. Find a particular solution, which depends on the right hand side. For a right hand side of the form
Pn(x) × exp{kx} (Pn(x) is a n-th order polynomial of x)
A particular solution is
yp = An(x) × exp{kx} = (a0 + a1x + a2x² + ... +anx^n) × exp{kx}
Problems arise if exp{kx} is a homogeneous solution of the DE, like in the given DE. In this case chose a polynomial of higher order than in the right hand side function:
yp = A(n+1)(x) × exp{kx}
Hence:
yp = (a0 + a1x + a2x²) × exp{2x}
yp' = [(2a0+a1) + 2(a1+a2)x + 2a2x² ] × exp{2x}
yp'' = [(4a0+4a1+2a2) + (4a1+8a2)x + 4a2x²] × exp{2x}
yp'' - 4yp = [(4a1+2a2) + 8a2x] × exp{2x}
Compare with the right hand side
[(4a1+2a2) + 8a2x] × exp{2x} = (8x - 2) × exp{2x}
and get the equation system
4a1 + 2a2 = -2
8 a2 = 8
with the solution
a1 = -1
a2 = 1
a0 remains as unknown constant (don't worry about this)
Therefore
yp = (x²-x+a0) × exp{2x}

3. sum up particular and homogeneous solution
y = yp +yh = c1 × exp{-2x} + (x²-x+a0+c2) × exp{2x}
substitute the sum of to unknown constants by one
c3 = a0 + c2
y = yp +yh = c1 × exp{-2x} + (x²-x+c3) × exp{2x}

Answer 2

Set y = Axe^(2x) + Be^(2x).

I'm pretty sure that will give you two equations in two variables for A and B.

If not, try something else similar.

Answer 3

Ah that’s you again!
♠ y’’–4y= (8x -2)(e^(2x)); first I’d rather brush it again, so that
y=w(x)*exp(2x), hence
y’ = w’*exp(2x) +2w*exp(2x), hence
y’’ = w’’*exp(2x) +4w’*exp(2x) +4w*exp(2x); thus
y’’-4y = (w’’ +4w’ +4w –4w)*exp(2x); and
♣ w’’+4w’ = 8x-2; or w’ +4w =4x^2-2x +C; now we may consider w’+4w =0, hence w=p*exp(-4x), where p is integration constant here, as well as variable to be found;
♦ w’ = p’*exp(-4x) –4p*esp(-4x); w’ +4w= p’*exp(-4x) =4x-2x;
p’=4x^2*exp(4x) –2x*exp(4x) +C*exp(4x); and
p= 4s1-2s2 +C1*exp(4x);
♥ now s1=∫x^2*exp(4x)=
= {by parts: u=x^2, du=2xdx, dv=exp(4x)*dx, v=0.25*exp(4x)} =
= 0.25x^2*exp(4x) -0.5∫x*exp(4x)*dx =0.25x^2*exp(4x) -0.5*s2;
s2=∫ x*exp(4x)*dx = {by parts} = 0.25x*exp(4x) –0.0625*exp(4x) ;
☻ (p-D)*exp(-4x) =4(0.25x^2 -0.5*(0.25x –0.0625)) –2(0.25x-0.0625) +C1;
(p-D)*exp(-4x) = x^2 –x+C2; p=(x^2 –x+C2)*exp(4x) +D;
w=(x^2 –x+C2) +D*exp(-4x); y=(x^2 –x+C2)*exp(2x) +D*exp(-2x);
☼ yes, schmiso and me coincide!