Solve the initial-value problem. Select the correct answer.
$$r ^ { t } + 2 t r = r , \quad r ( 0 ) = 2$$
A) $r ( t ) = 2 e ^ { t - t ^ { 2 } }$
B) $r ( t ) = 2 e ^ { t - 2 t }$
C) $r ( t ) = e ^ { t - 2 t ^ { 2 } }$
D) $r ( t ) = \frac { e ^ { t ^ { 2 } } } { 2 }$
E) $r ( t ) = 2 e ^ { 2 t }$

Question

Quizplus · Accepted Answer

Explanation:Solve the equation by using the integrating factor.$r ^ { t } + 2 t r = r$$\Rightarrow r ^ { t } + 2 t r - r = 0$$\Rightarrow r ^ { t } + ( 2 t - 1 ) r = 0$The integrating factor is$e ^ { \int ( 2 t - 1 ) d t } = e ^ { t ^ { 2 } - t }$Multiplying both sides by the integrating factor, we get$e ^ { t ^ { 2 } - t } r ^ { t } + ( 2 t - 1 ) e ^ { t ^ { 2 } - t } r = 0$$\frac { d } { d t } ( e ^ { t ^ { 2 } - t } r ) = 0$$e ^ { t ^ { 2 } - t } r = C$$r = C e ^ { - t ^ { 2 } + t }$Using the initial condition $r ( 0 ) = 2$, we get$2 = C e ^ { - 0 ^ { 2 } + 0 }$$C = 2$Therefore, the solution is$r ( t ) = 2 e ^ { - t ^ { 2 } + t } = 2 e ^ { t - t ^ { 2 } }$

Solve the Initial-Value Problem rt+2tr=r,r(0)=2r ^ { t } + 2 t r = r , \quad r ( 0 ) = 2rt+2tr=r,r(0)=2

Solve the Initial-Value Problem $r ^ { t } + 2 t r = r , \quad r ( 0 ) = 2$