Question

Solve the given differential equation by undetermined coefficients. y'' − y' − 12y = e4x

Answer 1

The general solution of the given differential equation be $y(x)=c_1e^{4x}+c_2e^{-3x}+\frac{1}{7} xe^{4x}$

For given question,

We have been given a differential equation y'' − y' − 12y = e^(4x)

We need to solve the given differential equation by undetermined coefficients.

We can write given differential equation as $(D^2-D-12)y=e^{4x}$ where $D=\frac{d}{dx}$

First solve the corresponding homogeneous differential equation:

y'' - y' - 12y = 0

The characteristic equation is:

m² - m - 12 =0

Let's find the roots of above quadratic equation by factorization.

⇒ m² - m - 12 = 0

⇒ (m - 4)(m + 3) = 0

⇒ m - 4 = 0   OR  m + 3 = 0

⇒ m = 4   OR   m = -3

Hence the complementary solution is,

$y_c=C_1e^{4x}+C_2e^{-3x}$

Let the general solution of a second order homogeneous differential equation be $y(x)=C_1(x)e^{4x}+C_2(x)e^{-3x}$

The unknown functions C1(x) and C2(x) can be determined from the system of two equations:

$C'_1e^{4x}+C'_2e^{-3x}=0~~~~~~~............(1)\\\\C'_1(4e^{4x})+C'_2(-3e^{-3x})=e^{4x}~~~~~~~...................(2)$

from (1),

$C'_2e^{-3x}=-C'_1e^{4x}$

Substitute this value in equation (2)

$4C'_1e^{4x}+3C'_1e^{4x}=e^{4x}$

$\Rightarrow C'_1=\frac{1}{7}$  and  $C'_2=-\frac{1}{7}e^{7x}$

$\Rightarrow C_1=\frac{1}{7} x+c_3$     and     $C_2=\frac{1}{49}e^{7x}+c_2$

Therefore, the general solution of the given differential equation be $y(x)=c_1e^{4x}+c_2e^{-3x}+\frac{1}{7} xe^{4x}$

Learn more about the differential equation here:

brainly.com/question/13234334

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