Solve the differential equation. $$\frac { d y } { d t } = - 12 t ^ { 2 } , \quad y ( 0 ) = 5$$
A) $y ( t ) = - 4 t ^ { 3 } - 5$
B) $y ( t ) = - 4 t ^ { 3 } + 5$
C) $y ( t ) = - 4 t ^ { 3 } + 1$
D) $y ( t ) = - 4 t ^ { 3 } + C$
E) $y ( t ) = - 24 t + 5$

Question

Solve the differential equation. $$\frac { d y } { d t } = - 12 t ^ { 2 } , \quad y ( 0 ) = 5$$ 
A) $y ( t ) = - 4 t ^ { 3 } - 5$
B) $y ( t ) = - 4 t ^ { 3 } + 5$
C) $y ( t ) = - 4 t ^ { 3 } + 1$
D) $y ( t ) = - 4 t ^ { 3 } + C$
E) $y ( t ) = - 24 t + 5$

Quizplus · Accepted Answer

To solve the differential equation, integrate $-12t^2$ with respect to $t$, which gives $-4t^3 + C$. Using the initial condition $y(0) = 5$, we find $C = 5$, so the solution is $y(t) = -4t^3 + 5$.

Solve the Differential Equation dydt=−12t2,y(0)=5\frac { d y } { d t } = - 12 t ^ { 2 } , \quad y ( 0 ) = 5dtdy​=−12t2,y(0)=5

Solve the Differential Equation $\frac { d y } { d t } = - 12 t ^ { 2 } , \quad y ( 0 ) = 5$