Question

Solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{16}\mathit{y}\mathbf{=}\mathbf{2}{\mathit{e}}^{\mathbf{4}\mathbf{x}}$

Short answer

The general solution of the given differential equation is

$y=c_1{e}^{-4x}+c_2{e}^{4x}+\frac{1}{4}x{e}^{4x}$

Step-by-step solution

Note the given data

Given the second order differential equation $y\text{'}\text{'}-16y=2e^{4x}$.

Obtain the general solution of the homogeneous differential equation

We know a homogeneous differential equation is an equation containing

differentiation and a function with a set of variables.

First, we need to solve the associated homogeneous equation $y\text{'}\text{'}-16y=0$

Corresponding auxiliary equation is l.

Solving,

m₁ =-4 and m₂ = 4, which are real and distinct.

So the complementary function is

$y_c=c_1{e}^{-4x}+c_2{e}^{4x}$

Finding the particular solution

Since the input function is a polynomial, the particular solution is also in the form of quadratic polynomial.

Particular solution will be $y_p=Ax{e}^{4x}$……(1)

We can find the coefficient A for which y_p is a solution of the differential equation.

Differentiate with respect to x

$y_p\text{'}=(4x+1)A{e}^{4x}$

$y_p\text{'}\text{'}=(16x+8)A{e}^{4x}$……(2)

Substituting (1) and (2) in given differential equation

$(16x+8)A{e}^{4x}-16Ax{e}^{4x}=2e^{4x}$

$8A{e}^{4x}=2e^{4x}$

comparing the coefficients,

8A =2$\Rightarrow A=\frac{1}{4}$

$-10A=20\Rightarrow A=-2$

Hence the particular solution is

1. becomes

$y_p=\frac{1}{4}x{e}^{4x}$

Finding the solution

Hence the solution of given differential equation is

y=y_c+y_p