Answer:
This differential equation can be solved with the method of the undetermined coefficients
The general solution is yg = yh + yp
where yh is the solution to the homogeneous equation and yp is a particular solution
first of all, we calculate the homogeneous solution
y'' + 36y = 0
with the characteristic polynomial
[tex]m^{2} - 36 = 0[/tex]
m = ± 6i
here we have imaginary solutions, hence:
yh = C1*cos(6t) + C2*sin(6t)
With C1 and C2 constants that we have to compute with the initial conditions. For the calculation of yp we assume that one solution is yp = C3*e^(-t)
by calculating the derivatives and replacing:
yp' = -C3*e^(-t)
yp'' = C3*e^(-t)
yp'' + 36yp = e^(-t)
C3*e^(-t) + 36*C3*e^(-t) = e^(-t)
37C3 = 1 --> C3 = 1/37
Hence, the solution is
y = yg = yh + yp = C1*cos(6t) + C2*sin(6t) + (1/37)e^(-t)
By using the initial condition we have
y(0) = C1*cos(0) + C2*sin(0) + (1/37)e^(0) = C1 + 1/37 = 0 --> C1 = -1/37
y'(0) = -C1*sin(0) + C2*cos(0) - (1/37)e^(0) = C2 - 1/37 = 0 --> C2= 1/37
Thus, finally
y = (1/37)( cos(6t) + sin(6t) + e^(-t) )
The final solution of the equation will be
[tex]Y=\dfrac{1}{37} (Cos(6t)+Sin(6t)+e^{-t}[/tex]
We will solve the differential equation by the method of undetermined coefficients.
The general solution will be
[tex]Y_g=Y_h+Y_p[/tex]
Here [tex]Y_h[/tex] is the solution of the given homogeneous equation and [tex]Y_p[/tex] is a particular solution
first, we will calculate the homogeneous solution
[tex]Y''+36y=0[/tex]
The characteristic polynomial
[tex]m^2 -36=0[/tex]
[tex]m=\pm6i[/tex]
Since we are having imaginary solution of the equation
[tex]Y_h= C_1cos(6t)+C_2sin(6t)[/tex]
Here [tex]C_1[/tex] and [tex]C_2[/tex] are constants. Now for the calculation [tex]Y_p[/tex] we assume that one solution will be
[tex]Y_p=C_3\times e^{-t}[/tex]
From taking the derivative
[tex]Y_p =-C_3e^{-t}\\Y''_p=C_3e^{-t}\\Y''_p+36Y_p=-e^{-t}\\C_3e^{-t}+36C_3e^{-t}=e^{-t}[/tex]
[tex]37\ C_3=1[/tex]
[tex]C_3=\dfrac{1}{37}[/tex]
Hence the solution will be
[tex]Y_g=Y_h+Y_p[/tex]
[tex]C_1Cos(6t)+C_2sin(6t)+\dfrac{1}{37}e^(-t)[/tex]
Now from initial conditions
[tex]Y(0)=C_1Cos(0)+C_2sin(0)+\dfrac{1}{37} e^(0)=C_1+\dfrac{1}{37}=0=C_1=\dfrac{-1}{37}\\Y'(0)=-C_1sin(0)+C_2cos(0)+\dfrac{1}{37} e^(0)=C_2-\dfrac{1}{37}=0=C_1=\dfrac{1}{37}[/tex]
Thus Finally
[tex]Y=\dfrac{1}{37} (cos(6t)+sin(6t)+e^-t )[/tex]
Thus the solution of the given differential equation will be
[tex]Y=\dfrac{1}{37} (cos(6t)+sin(6t)+e^-t )[/tex]
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