# Math Help - Particular Solution....

1. ## Particular solution 2

d^2x/[dt]^2 -4 dx/dt=7-3e^4t

Could someone also help me with this? Find the particular solution. I find trouble knowing what trial to use because of the 7-3e^4t. My Tutor said to use a constant with the particular trial >>xp(t)=Ate^4t. Can you show me how to get the A,B and constant value please. Thanx

2. Let $x_p(t) = Ate^{4t}$ then

$\frac{dx}{dt} = 4Ate^{4t} + Ae^{4t}$ and

$\frac{d^2x}{dt^2} = 4Ae^{4t} + 16Ate^{4t} + 4Ae^{4t}$

then you have

$\frac{d^2x}{dt^2} - 4\frac{dx}{dt} = 4Ae^{4t} + 16Ate^{4t} + 4Ae^{4t} -4(4Ate^{4t} + Ae^{4t})$ = $4Ae^{4t} = -3e^{4t}$, so $A = -\frac{3}{4}$. Hence $x_p(t) = -\frac{3}{4}te^{4t}$.

For the complementary solution try $x_c(t) = Bt$.

Then $-4\frac{dx}{dt} = -4B = 7$ so $B = -\frac{7}{4}$.

Hence $x(t) = x_p(t) + x_c(t) = -\frac{3}{4}te^{4t} - \frac{7}{4}t$.