Answer to Question #110911 in Math for sebastian staz

Question

y(t) + 4 y(t) = 6 e^2t with y(0) = 3 find solution of this differential equation with use laplace transform.

Answer 1

In order to solve differential equation using Laplace transform, one has to take Laplace transform of both sides of equation, using the properties of the transform, solve for image $Y(s) \equiv L[y(t)]$, and take inverse Laplace transform.

Using property $L[y'(t)] = s Y(s) - y(0)$, table Laplace transform $L(e^{a t}) = \frac{1}{s-a}$, and taking transform of both sides of $y'(t) + 4 y(t) = 6 e^{2 t}$, obtain:

$s Y(s) - y(0) + 4 Y(s) = \frac{6}{s-2}$.

Using initial condition $y(0) = 3$ and simplifying, obtain $Y(s)[s + 4] = \frac{6}{s-2} + 3 = \frac{3 s}{s-2}$, from where $Y(s) = \frac{3 s}{(s+4)(s-2)}$.

In order to take inverse Laplace transform, one has to expand the last expression into partial fractions: $\frac{3 s}{(s+4)(s-2)} = \frac{A}{s+4} + \frac{B}{s-2}$. Putting terms in the right hand side to common denominator, and equating the coefficients next to powers of $s$ in the numerator of left and right hand side, obtain system of linear equations for $A, B$:

$A+B = 3; -2 A + 4 B = 0$, which has roots $A = 2; B = 1$.

Hence, $\frac{3 s}{(s+4)(s-2)} = \frac{2}{s+4} + \frac{1}{s-2}$.

From table transform $L[e^{a t}] = \frac{1}{s-a}$, the inverse of expressions like $\frac{1}{s-a}$ is $L^{-1}[\frac{1}{s-a}] = e^{a t}$.

Therefore, the solution is $y(t) = L^{-1}[Y(s)] = L^{-1}[\frac{2}{s+4} + \frac{1}{s-2}] = 2 e^{-4 t} + e^{2 t}$.