Solve the equation using an integrating factor.
-$\mathrm { y } ^ { \prime } + \frac { 4 } { \mathrm { t } } \mathrm { y } = \mathrm { t } ^ { 4 }$ , t 0
A) y = C $t ^ { 4 }$ + $t ^ { 5}$
B) y = - $t ^ { - 4 }$ + $t ^ { 3 }$ + C
C) y = $t ^ { - 4 }$ + $t ^ { 5 }$
D) y = C $t ^ { - 4 }$ + $\frac { 1 } { 9 }$ $t ^ { 5 }$

Question

Solve the equation using an integrating factor. 
-$\mathrm { y } ^ { \prime } + \frac { 4 } { \mathrm { t } } \mathrm { y } = \mathrm { t } ^ { 4 }$ , t > 0
A) y = C $t ^ { 4 }$ + $t ^ { 5}$
B) y = - $t ^ { - 4 }$ + $t ^ { 3 }$ + C
C) y = $t ^ { - 4 }$ + $t ^ { 5 }$
D) y = C $t ^ { - 4 }$ + $\frac { 1 } { 9 }$ $t ^ { 5 }$

Quizplus · Accepted Answer

The integrating factor is $e^{\int \frac{4}{t} dt} = e^{4\ln t} = t^4$. Multiplying both sides by $t^4$ gives $t^4y' + 4t^3y = t^8$. Integrating $\int (t^4y)' dt = \int t^8 dt$ gives $t^4y = \frac{1}{9}t^9 + C$, so $y = C t^{-4} + \frac{1}{9}t^5$.

Solve the Equation Using an Integrating Factor y′+4ty=t4\mathrm { y } ^ { \prime } + \frac { 4 } { \mathrm { t } } \mathrm { y } = \mathrm { t } ^ { 4 }y′+t4​y=t4

Solve the Equation Using an Integrating Factor $\mathrm { y } ^ { \prime } + \frac { 4 } { \mathrm { t } } \mathrm { y } = \mathrm { t } ^ { 4 }$