Solve the initial value problem.
-$$\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { d } \mathrm { t } ^ { 2 } } = \mathrm { e } ^ { 2 \mathrm { t } } + 10 \sin \mathrm { t } , \mathrm { y } ( 0 ) = 0 , \mathrm { y } ^ { \prime } ( 0 ) = 6$$
A) $y = \frac { e ^ { 2 t } } { 4 } - 10 \sin t$
B) $y = \frac { e ^ { 2 t } } { 4 } - 10 \sin t + 6 t - \frac { 1 } { 4 }$
C) $y = e ^ { 2 t } - 10 \sin t + 15 t - \frac { 1 } { 4 }$
D) $y = \frac { e ^ { 2 t } } { 4 } - 10 \sin t + \frac { 31 } { 2 } t - \frac { 1 } { 4 }$

Question

Quizplus · Accepted Answer

The given differential equation is a second-order linear non-homogeneous differential equation.

First, we find the general solution of the homogeneous equation:
$$\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { d } \mathrm { t } ^ { 2 } } = 0$$

The characteristic equation is: 
$$r^2 = 0 \implies r = 0,0$$

Therefore, the general solution of the homogeneous equation is: 
$$y_h = c_1 + c_2 t$$

Now, we find the particular solution of the non-homogeneous equation using the method of undetermined coefficients.

For the non-homogeneous part: 
$$f(t) = e^{2t} + 10\sin t$$

We assume the particular solution of the form: 
$$y_p = A e^{2t} + B \cos t + C \sin t$$ 
where A, B, and C are constants to be determined.

Then, we have: 
$$\frac { d ^{2} y_p } { d t^2 } = 4A e^{2t} - B \cos t - C \sin t$$

Substituting the particular solution and its derivatives into the differential equation, we get: 
$$4A e^{2t} - B \cos t - C \sin t = e^{2t} + 10\sin t$$

Comparing the coefficients of like terms, we get: 
$$4A = 1 \implies A = \frac {1}{4}$$ 
$$B = 0$$ 
$$C = -10$$

Therefore, the particular solution of the non-homogeneous equation is: 
$$y_p = \frac { e^{2t} }{ 4 } - 10 \sin t$$

Hence, the general solution of the non-homogeneous equation is: 
$$y = y_h + y_p = c_1 + c_2 t + \frac { e^{2t} }{ 4 } - 10 \sin t$$

Using the initial conditions: 
$$y(0) = 0 \implies c_1 = -\frac{1}{4}$$ 
$$y'(0) = 6 \implies c_2 = \frac{31}{2}$$

Therefore, the solution of the initial value problem is: 
$$y = -\frac{1}{4} + \frac { e^{2t} }{ 4 } - 10 \sin t + \frac{31}{2} t$$

which matches exactly with option D.

Solve the Initial Value Problem A) y=e2t4−10sin⁡ty = \frac { e ^ { 2 t } } { 4 } - 10 \sin ty=4e2t​−10sint

Solve the Initial Value Problem A) $y = \frac { e ^ { 2 t } } { 4 } - 10 \sin t$