Solve the initial value problem.
-$$\frac { d ^ { 2 } y } { d x ^ { 2 } } = 5 - 3 x , y ^ { \prime } ( 0 ) = 8 , y ( 0 ) = 2$$
A) $y = 5 x ^ { 2 } + 3 x ^ { 3 } + 8 x + 2$
B) $y = \frac { 5 } { 2 } x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 } - 8 x - 2$
C) $\mathrm { y } = 2$
D) $y = \frac { 5 } { 2 } x ^ { 2 } - \frac { 1 } { 2 } x ^ { 3 } + 8 x + 2$

Question

Quizplus · Accepted Answer

The characteristic equation for the differential equation is $r^2 = -3r+5$, or equivalently, $r^2 + 3r - 5 = 0$. Factoring the quadratic gives $(r+5)(r-2) = 0$, so the roots are $r=-5$ and $r=2$. Thus, the general solution to the differential equation is $y(x) = c_1e^{-5x} + c_2e^{2x} + \frac{5}{6}-\frac{3}{4}x$, where $c_1$ and $c_2$ are constants determined by the initial conditions $y(0) = 2$ and $y'(0) = 8$.

Using the initial condition $y(0) = 2$, we see that $\frac{5}{6} = 2$, which is not true. So we can eliminate choices C and A.

Differentiating the general solution, we have $y'(x) = -5c_1e^{-5x} + 2c_2e^{2x} - \frac{3}{4}$. Using the initial condition $y'(0) = 8$, we can solve for $c_1$ and $c_2$:
\begin{align*}
8 &= y'(0) = -5c_1 + 2c_2 - \frac{3}{4} \
\Rightarrow c_1 &= \frac{23}{20} - \frac{2}{5}c_2
\end{align*}

Substituting this into the general solution and using the initial condition $y(0) = 2$, we get:
\begin{align*}
2 &= y(0) = c_1 + c_2 + \frac{5}{6} \
&= \left(\frac{23}{20}-\frac{2}{5}c_2\right) + c_2 + \frac{5}{6} \
\Rightarrow c_2 &= -\frac{11}{10}
\end{align*}

Therefore, the particular solution to the differential equation that satisfies the initial conditions is
$$y(x) = \frac{5}{2}x^2 - \frac{1}{2}x^3 + 8x + 2,$$
so the answer is $\boxed{\textbf{(D)}}$.

Solve the Initial Value Problem d2ydx2=5−3x,y′(0)=8,y(0)=2\frac { d ^ { 2 } y } { d x ^ { 2 } } = 5 - 3 x , y ^ { \prime } ( 0 ) = 8 , y ( 0 ) = 2dx2d2y​=5−3x,y′(0)=8,y(0)=2

Solve the Initial Value Problem $\frac { d ^ { 2 } y } { d x ^ { 2 } } = 5 - 3 x , y ^ { \prime } ( 0 ) = 8 , y ( 0 ) = 2$