Answer to Question #127081 in Differential Equations for jse

Question

Use the Laplace transform to solve the given initial value problem.

y′′−4y′−96y=0; y(0)=14, y′(0)= 28


Enclose arguments of functions in parentheses. For example, sin(2x).

y =___________

Answer 1

Solution:

$y(x)\leftrightarrow Y(p)$

$y'(x)\leftrightarrow pY(p)-y(0)=pY(p)-14$

$y''(x)\leftrightarrow p^2Y(p)-py(0)-y'(0)=p^2Y(p)-14p-28$

We compose an operator equation that corresponds to the given differential equation:

$p^2Y(p)-14p-28-4(pY(p)-14)-96Y(p)=0$

$(p^2-4p-96)Y(p)=14p-28$

$Y(p)=(14p-28)/(p^2-4p-96)$

$Y(p)=(14p-28)/((p+8)(p-12))$

$y(x)=Res_{p=-8}(Y(p)e^{px})+Res_{p=12}(Y(p)e^{px})=$

$=\lim\limits_{p\rarr-8}\frac{14p-28}{p-12}e^{px}+\lim\limits_{p\rarr12}\frac{14p-28}{p+8}e^{px}=$

$=\frac{-140}{-20}e^{-8x}+\frac{140}{20}e^{12x}=7e^{-8x}+7e^{12x}$

Answer: $y(x)=7e^{(-8x)}+7e^{(12x)}$