Solve the initial value problem.
-$$t \frac { d y } { d t } + 4 y = t ^ { 3 } ; t 0 , y ( 2 ) = 1$$
A) $y = \frac { t ^ { 3 } } { 7 } + \frac { 144 } { 7 } t ^ { - 4 } , t 0$
B) $y = \frac { t ^ { 3 } } { 7 } - \frac { 16 } { 7 } t ^ { - 4 } , t 0$
C) $y = \frac { t ^ { 3 } } { 7 } - 2 t ^ { - 4 } , t 0$
D) $y = \frac { t ^ { 3 } } { 7 } + \frac { 16 } { 7 } t ^ { - 4 } , t 0$

Question

Solve the initial value problem.
-$$t \frac { d y } { d t } + 4 y = t ^ { 3 } ; t > 0 , y ( 2 ) = 1$$
A) $y = \frac { t ^ { 3 } } { 7 } + \frac { 144 } { 7 } t ^ { - 4 } , t > 0$
B) $y = \frac { t ^ { 3 } } { 7 } - \frac { 16 } { 7 } t ^ { - 4 } , t > 0$
C) $y = \frac { t ^ { 3 } } { 7 } - 2 t ^ { - 4 } , t > 0$
D) $y = \frac { t ^ { 3 } } { 7 } + \frac { 16 } { 7 } t ^ { - 4 } , t > 0$

Quizplus · Accepted Answer

The given differential equation is a linear first-order equation. Using an integrating factor, we find the general solution $y = \frac{t^3}{7} + Ct^{-4}$. Applying the initial condition $y(2) = 1$, we solve for $C$ and find $C = -\frac{16}{7}$. Thus, the solution is $y = \frac{t^3}{7} - \frac{16}{7}t^{-4}$.

Solve the Initial Value Problem tdydt+4y=t3;t>0,y(2)=1t \frac { d y } { d t } + 4 y = t ^ { 3 } ; t > 0 , y ( 2 ) = 1tdtdy​+4y=t3;t>0,y(2)=1

Solve the Initial Value Problem $t \frac { d y } { d t } + 4 y = t ^ { 3 } ; t > 0 , y ( 2 ) = 1$