# Question 95164

May 4, 2017

(d^2y)/(dt^2)− y = 2e^t qquad star

The general solution (${y}_{g}$) is made up of the complementary (${y}_{c}$) and particular (${y}_{p}$) solutions.

The complementary solution is the solution to homogeneous DE:

(d^2y)/(dt^2)− y = 0#

This has characteristic equation ${\lambda}^{2} - 1 = 0$, $\lambda = \pm 1$

And so ${y}_{c} = A {e}^{t} + B {e}^{- t}$

For the particular solution, we might think that ${y}_{g} = \alpha {e}^{t}$ is the way to go but that term is already in the complementary solution and so we look instead at ${y}_{p} = \alpha t {e}^{t}$

So:

${y}_{p} = \alpha t {e}^{t}$

${y}_{p} ' = \alpha {e}^{t} + \alpha t {e}^{t} = \alpha {e}^{t} \left(1 + t\right)$

${y}_{p} ' ' = \alpha {e}^{t} \left(1 + t\right) + \alpha {e}^{t} = \alpha {e}^{t} \left(2 + t\right)$

Plugging that into $\star$ yields:

$\alpha {e}^{t} \left(2 + t\right) - \alpha t {e}^{t} = 2 {e}^{t} \implies \alpha = 1$

So the general solution is:

${y}_{g} = A {e}^{t} + B {e}^{- t} + t {e}^{t}$

May 4, 2017

$y \left(t\right) = t {e}^{t} + {c}_{1} {e}^{t} + {c}_{2} {e}^{- t}$

#### Explanation:

First, solve for $\frac{{d}^{2} y}{\mathrm{dt}} ^ 2 - y = 0$ to obtain the solution of the associated homogeneous equation. This gives ${c}_{1} {e}^{t} + {c}_{2} {e}^{- t}$.

Now try to solve $\frac{{d}^{2} y}{\mathrm{dt}} ^ 2 - y = 2 {e}^{t}$. Guess that a particular solution $y$ is of the form $A {e}^{t}$, where $A$ is a constant. Thus, $A {e}^{t} - A {e}^{t} = 2 {e}^{t}$, which is clearly not possible. Modify our guess to $A t {e}^{t}$. Thus, $A \left(t {e}^{t} + {e}^{t}\right) + A {e}^{t} - A t {e}^{t} = 2 {e}^{t}$, or $A = 1$. This means that $t {e}^{t}$ is one of the solutions. Now, using the solutions of the associated homogenous equation, all solutions $y \left(t\right)$ of this differential equation will satisfy $y \left(t\right) - t {e}^{t} = {c}_{1} {e}^{t} + {c}_{2} {e}^{- t}$. Thus, all solutions satisfy $y \left(t\right) = t {e}^{t} + {c}_{1} {e}^{t} + {c}_{2} {e}^{- t}$.