Solve the initial value problem for x as a function of t.
-$$\left( t ^ { 2 } - 5 t + 6 \right) \frac { d x } { d t } = 1 \quad ( t 3 ) , x ( 4 ) = 0$$
A) $x = - \ln | t - 2 | + \ln | t - 3 | + \ln 2$
B) $x = - \ln | t - 2 | + \ln | t - 3 | + 2$
C) $x = - \ln | t - 2 | + \frac { 1 } { t - 3 } + \ln 2$
D) $x = \ln | t - 2 | - \ln | t - 3 | + \ln 2$

Question

Solve the initial value problem for x as a function of t.
-$$\left( t ^ { 2 } - 5 t + 6 \right) \frac { d x } { d t } = 1 \quad ( t > 3 ) , x ( 4 ) = 0$$
A) $x = - \ln | t - 2 | + \ln | t - 3 | + \ln 2$
B) $x = - \ln | t - 2 | + \ln | t - 3 | + 2$
C) $x = - \ln | t - 2 | + \frac { 1 } { t - 3 } + \ln 2$
D) $x = \ln | t - 2 | - \ln | t - 3 | + \ln 2$

Quizplus · Accepted Answer

To solve the initial value problem, first, integrate both sides with respect to $t$. The left side involves the integral of $\frac{1}{t^2 - 5t + 6}$, which can be factored as $\frac{1}{(t-2)(t-3)}$. Using partial fraction decomposition, this can be rewritten and integrated to yield $-\ln|t-2| + \ln|t-3|$. Applying the initial condition $x(4) = 0$ allows for solving the constant of integration, resulting in the solution $x = - \ln | t - 2 | + \ln | t - 3 | + \ln 2$.

Solve the Initial Value Problem for X as a Function (t2−5t+6)dxdt=1(t>3),x(4)=0\left( t ^ { 2 } - 5 t + 6 \right) \frac { d x } { d t } = 1 \quad ( t > 3 ) , x ( 4 ) = 0(t2−5t+6)dtdx​=1(t>3),x(4)=0

Solve the Initial Value Problem for X as a Function $\left( t ^ { 2 } - 5 t + 6 \right) \frac { d x } { d t } = 1 \quad ( t > 3 ) , x ( 4 ) = 0$