# Find the general solution for \ddot{y} + 5\dot{y} + 4y= 16t + 12e^{-4t} + 20e^{-6t}.

## Question:

Find the general solution for {eq}\ddot{y} + 5\dot{y} + 4y= 16t + 12e^{-4t} + 20e^{-6t}. {/eq}

## Second Order Linear Differential Equations.

Second Order Linear Differential Equations are equations of the form:

{eq}p(t) y '' + q(t) y' + h(t) = g(t) {/eq}

When the function {eq}g(t) = 0 {/eq} we call the equation and homogeneous differential equation.

The characteristic equation is: {eq}p(t) r^2 + q(t) r + h(t) = 0 {/eq}

Solving this quadratic equation we get two solutions : {eq}r_1 , r_2 {/eq}

And then the general solution for the homogeneous equation would be:

{eq}y = c_1 e^{r_1t} + c_2 e^{r_2t} {/eq}

## Answer and Explanation:

The corresponding homogeneous equation is:

{eq}\ddot{y} + 5\dot{y} + 4y= 0 {/eq}

Step 1: Find and solve the characteristic equation:

The characteristic equation is {eq}r^2 + 5r + 4 =0 {/eq}

Solving it we get:

{eq}(r+4) (r+1 ) = 0 \\ \Rightarrow r_1=-1 , r_2=-4 {/eq}

Step 2: Find the general solution:

According to the general solution formula for second order linear equations:

{eq}y = c_1 e^{r_1t} + c_2 e^{r_2t} = c_1 e^{-t} + c_2 e^{-4t} {/eq}

#### Learn more about this topic:

Separable Differential Equation: Definition & Examples

from GRE Math: Study Guide & Test Prep

Chapter 16 / Lesson 1
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