Solve the initial-value problem $\frac { d y } { d t } = 2 t y ^ { 2 } + 3 y ^ { 2 }$ , $y ( 0 ) = 1$ . Then use your solution to evaluate $y ( 1 )$ .
A) $- 3$
B)1
C) $- \frac { 1 } { 3 }$
D) $\frac { 1 } { 3 }$
E)0
F)3

G) $- \frac { 1 } { 5 }$
H) $\frac { 1 } { 5 }$

Question

Solve the initial-value problem $\frac { d y } { d t } = 2 t y ^ { 2 } + 3 y ^ { 2 }$ , $y ( 0 ) = 1$ . Then use your solution to evaluate $y ( 1 )$ .
A) $- 3$ 
B)1
C) $- \frac { 1 } { 3 }$ 
D) $\frac { 1 } { 3 }$ 
E)0
F)3

G) $- \frac { 1 } { 5 }$ 
H) $\frac { 1 } { 5 }$

Quizplus · Accepted Answer

The differential equation is separable. Separating variables and integrating, we find the solution $y(t) = \frac{1}{1 - t^2 - 3t}$. Evaluating at $t = 1$, $y(1) = \frac{1}{1 - 1 - 3} = -\frac{1}{3}$.

Solve the Initial-Value Problem dydt=2ty2+3y2\frac { d y } { d t } = 2 t y ^ { 2 } + 3 y ^ { 2 }dtdy​=2ty2+3y2

Solve the Initial-Value Problem $\frac { d y } { d t } = 2 t y ^ { 2 } + 3 y ^ { 2 }$