Solve the differential equation.
-$$\frac { d y } { d x } = 4 x ^ { 3 } \sqrt { y - 1 }$$
A) $y = \frac { 1 } { 4 } x ^ { 8 } + C$
B) $y = \left( 2 x ^ { 4 } + C \right) ^ { 2 } + 1$
C) $y = \left( \frac { 1 } { 2 } x ^ { 4 } + C \right) ^ { 2 } + 1$
D) $y = \left( x ^ { 4 } + C \right) ^ { 2 }$

Question

Quizplus · Accepted Answer

Let us separate the variables and integrate both sides:
$$\int \frac{dy}{\sqrt{y-1}} = \int 4x^3\,dx$$
Making the substitution $y-1 = t^2$ (so $\frac{dy}{dt}=2t$), we get:
$$\int \frac{2t}{t}\,dt = \int 4x^3\,dx$$
which simplifies to:
$$\int 2\,dt = x^4 + C$$
Solving for $y$, we get:
$$y = t^2 + 1 = \left(\frac{1}{2} x^4 + C ight)^2 + 1$$
So the answer is $\boxed{	extbf{(C)}}$.

Solve the Differential Equation dydx=4x3y−1\frac { d y } { d x } = 4 x ^ { 3 } \sqrt { y - 1 }dxdy​=4x3y−1​

Solve the Differential Equation $\frac { d y } { d x } = 4 x ^ { 3 } \sqrt { y - 1 }$