Solve the initial value problem using an integrating factor.
-t $y ^ { \prime }$ + 3y = 5t; $y ( 1 ) = 1$ , t 0
A) y = $\frac { 5 } { 4 }$ t - $\frac { 1 } { 4 t ^ { 3 } }$
B) y = $\sqrt { \frac { 4 } { 3 } t ^ { 2 } }$ + 1 - $\sqrt { \frac { 4 } { 3 } }$
C) y = 5 $t ^ { 2 }$ - 4t
D) y = 5 $t ^ { 3 }$ - 4 $t ^ { 2 }$

Question

Solve the initial value problem using an integrating factor. 
-t $y ^ { \prime }$ + 3y = 5t; $y ( 1 ) = 1$ , t > 0
A) y = $\frac { 5 } { 4 }$ t - $\frac { 1 } { 4 t ^ { 3 } }$
B) y = $\sqrt { \frac { 4 } { 3 } t ^ { 2 } }$ + 1 - $\sqrt { \frac { 4 } { 3 } }$
C) y = 5 $t ^ { 2 }$ - 4t
D) y = 5 $t ^ { 3 }$ - 4 $t ^ { 2 }$

Quizplus · Accepted Answer

The given differential equation is a first-order linear differential equation. To solve it using an integrating factor, we first write it in standard form: $y' + \frac{3}{t}y = 5$. The integrating factor, $μ(t)$, is found by integrating the coefficient of $y$, which is $\frac{3}{t}$, giving $μ(t) = e^{\int \frac{3}{t} dt} = e^{3\ln t} = t^3$. Multiplying the entire equation by $t^3$ and integrating both sides with respect to $t$ allows us to find the general solution. Applying the initial condition $y(1) = 1$ helps to find the specific constant, leading to the solution $y = \frac{5}{4}t - \frac{1}{4t^3}$.

Solve the Initial Value Problem Using an Integrating Factor y′y ^ { \prime }y′

Solve the Initial Value Problem Using an Integrating Factor $y ^ { \prime }$