Solve the initial value problem for x as a function of t.
-$$\left( t ^ { 2 } + 2 t \right) \frac { d x } { d t } = 2 x + 8 , x ( 1 ) = 1$$
A) $x = 15 \ln \left| \frac { t } { t + 2 } \right| - 4$
B) $x = \frac { 14 t } { t + 2 } - 4$
C) $x = \frac { 15 t } { t + 2 } - 4$
D) $x = 15 t + \ln | t + 2 | - 4$

Question

Quizplus · Accepted Answer

We start by separating variables and integrating both sides: $\int \frac{2dx}{2x+8} = \int \frac{dt}{t^2+2t}$. Simplifying the left-hand side gives us $\ln|x+4|$. We can rewrite the right-hand side as $\int \frac{dt}{t(t+2)} = \frac{1}{2} \int \left(\frac{1}{t} - \frac{1}{t+2}\right) dt$, and simplify to $\frac{1}{2}\ln\left|\frac{t}{t+2}\right|$. Thus, we have $\ln|x+4| = \frac{1}{2}\ln\left|\frac{t}{t+2}\right| + C$ for some constant $C$. Using the initial condition $x(1) = 1$, we get $C = \ln(3) - \frac{1}{2}\ln(3) - \ln(5) + 2 = \ln\left|\frac{3}{5}\right| + 2$. Plugging in $C$ and isolating $x$, we get $x = \frac{15t}{t+2} - 4$. Therefore, the answer is choice C.

Solve the Initial Value Problem for X as a Function (t2+2t)dxdt=2x+8,x(1)=1\left( t ^ { 2 } + 2 t \right) \frac { d x } { d t } = 2 x + 8 , x ( 1 ) = 1(t2+2t)dtdx​=2x+8,x(1)=1

Solve the Initial Value Problem for X as a Function $\left( t ^ { 2 } + 2 t \right) \frac { d x } { d t } = 2 x + 8 , x ( 1 ) = 1$