Find the solution of the initial-value problem $\frac { d y } { d t } = 2 t \sqrt { 1 - y }$ , $y ( 1 ) = 0$ .
A) $2 \sqrt { y - 1 } = 3 - t ^ { 2 }$
B) $\frac { 2 } { \sqrt { y - 1 } } = 3 - t ^ { 2 }$
C) $2 \sqrt { 1 - y } = 3 + t ^ { 2 }$
D) $\frac { 2 } { \sqrt { 1 - y } } = 3 + t ^ { 2 }$
E) $2 \sqrt { 1 - y } = - 3 + t ^ { 2 }$
F) $\frac { 2 } { \sqrt { 1 - y } } = - 3 + t ^ { 2 }$
G) $2 \sqrt { 1 - y } = 3 - t ^ { 2 }$
H) $\frac { 2 } { \sqrt { 1 - y } } = 3 - t ^ { 2 }$

Question

Find the solution of the initial-value problem $\frac { d y } { d t } = 2 t \sqrt { 1 - y }$ , $y ( 1 ) = 0$ .
A) $2 \sqrt { y - 1 } = 3 - t ^ { 2 }$ 
B) $\frac { 2 } { \sqrt { y - 1 } } = 3 - t ^ { 2 }$ 
C) $2 \sqrt { 1 - y } = 3 + t ^ { 2 }$ 
D) $\frac { 2 } { \sqrt { 1 - y } } = 3 + t ^ { 2 }$ 
E) $2 \sqrt { 1 - y } = - 3 + t ^ { 2 }$ 
F) $\frac { 2 } { \sqrt { 1 - y } } = - 3 + t ^ { 2 }$ 
G) $2 \sqrt { 1 - y } = 3 - t ^ { 2 }$ 
H) $\frac { 2 } { \sqrt { 1 - y } } = 3 - t ^ { 2 }$

Quizplus · Accepted Answer

The differential equation can be solved by separation of variables. Integrating both sides and applying the initial condition $y(1) = 0$, we find that the solution is $2 \sqrt{1 - y} = 3 - t^2$.

Find the Solution of the Initial-Value Problem dydt=2t1−y\frac { d y } { d t } = 2 t \sqrt { 1 - y }dtdy​=2t1−y​

Find the Solution of the Initial-Value Problem $\frac { d y } { d t } = 2 t \sqrt { 1 - y }$