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Sagot :
The general solution of the given differential equation be [tex]y(x)=c_1e^{4x}+c_2e^{-3x}+\frac{1}{7} xe^{4x}[/tex]
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 [tex](D^2-D-12)y=e^{4x}[/tex] where [tex]D=\frac{d}{dx}[/tex]
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,
[tex]y_c=C_1e^{4x}+C_2e^{-3x}[/tex]
Let the general solution of a second order homogeneous differential equation be [tex]y(x)=C_1(x)e^{4x}+C_2(x)e^{-3x}[/tex]
The unknown functions C1(x) and C2(x) can be determined from the system of two equations:
[tex]C'_1e^{4x}+C'_2e^{-3x}=0~~~~~~~............(1)\\\\C'_1(4e^{4x})+C'_2(-3e^{-3x})=e^{4x}~~~~~~~...................(2)[/tex]
from (1),
[tex]C'_2e^{-3x}=-C'_1e^{4x}[/tex]
Substitute this value in equation (2)
[tex]4C'_1e^{4x}+3C'_1e^{4x}=e^{4x}[/tex]
[tex]\Rightarrow C'_1=\frac{1}{7}[/tex] and [tex]C'_2=-\frac{1}{7}e^{7x}[/tex]
[tex]\Rightarrow C_1=\frac{1}{7} x+c_3[/tex] and [tex]C_2=\frac{1}{49}e^{7x}+c_2[/tex]
Therefore, the general solution of the given differential equation be [tex]y(x)=c_1e^{4x}+c_2e^{-3x}+\frac{1}{7} xe^{4x}[/tex]
Learn more about the differential equation here:
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