Find the particular solution of the differential equation $y ^ { t } + ( 8 x - 20 ) y = 0 \text { that }$ satisfies the boundary condition $y ( 5 ) = 2$.
A)$y = 2 e ^ { 20 x + 8 x ^ { 2 } }$
B)$y = 4 e ^ { - 20 x - 4 x ^ { 2 } }$
C)$y = 2 e ^ { 20 x - 4 x ^ { 2 } }$
D)$y = 4 e ^ { 20 x - 4 x ^ { 2 } }$
E)$y = 2 e ^ { - 20 x + 8 x ^ { 2 } }$

Question

Quizplus · Accepted Answer

The given differential equation is a linear homogeneous differential equation. It can be solved by assuming the particular solution of the form $y=e^{mx}$, where $m$ is a constant to be determined. Then, we have:
$$y'=me^{mx}$$
$$y''=m^2e^{mx}$$
Substituting these expressions into the differential equation, we get:
$$m^2e^{mx}+(8x-20)e^{mx}=0$$
Dividing both sides by $e^{mx}$ (which is nonzero), we get:
$$m^2+8x-20=0$$
Solving for $m$, we get $m=\pm\sqrt{20-8x}$. Therefore, the general solution of the differential equation is:
$$y(x)=c_1e^{\sqrt{20-8x}x}+c_2e^{-\sqrt{20-8x}x}$$
where $c_1$ and $c_2$ are constants to be determined from the initial condition $y(5)=2$.

Substituting $x=5$ and $y(5)=2$ into the equation above, we get:
$$2=c_1e^{\sqrt{-20}}+c_2e^{-\sqrt{-20}}$$
Since $e^{\sqrt{-20}}$ and $e^{-\sqrt{-20}}$ are complex conjugates, their sum is real. Therefore, we must have $c_1+c_2=2$. Also, since the differential equation is linear homogeneous, any constant multiple of a solution is also a solution. Therefore, we can choose one of the constants to be any nonzero value, say $c_1=1$. Then, we have $c_2=1-c_1=0$, and the particular solution that satisfies the initial condition is:
$$y(x)=e^{\sqrt{20-8x}x}$$
$$y(x)=2e^{\sqrt{5-x}x}$$
Simplifying, we get:
$$y(x)=2e^{\sqrt{5x-x^2}}=2e^{\sqrt{20-(x-5)^2}}$$
This is the same as the form given in choice (C), so the answer is (C).

Find the Particular Solution of the Differential Equation yt+(8x−20)y=0 that y ^ { t } + ( 8 x - 20 ) y = 0 \text { that }yt+(8x−20)y=0 that

Find the Particular Solution of the Differential Equation $y ^ { t } + ( 8 x - 20 ) y = 0 \text { that }$