Find the general solution of the differential equation. $\frac { \mathrm { d } y } { \mathrm {~d} x } = ( x + 1 ) y ^ { 2 }$
Solve for y as a function of x.

A) $y = - ( x + 1 ) ^ { 2 } + C$
B) $y = - \frac { 2 } { x ^ { 2 } + C }$
C) $y = - \frac { 2 } { x ( x + 2 ) + C }$
D) $y = \frac { C } { ( x + 1 ) ^ { 2 } }$
E) $y = - \frac { 2 } { x + C }$

Question

Find the general solution of the differential equation.   $\frac { \mathrm { d } y } { \mathrm {~d} x } = ( x + 1 ) y ^ { 2 }$  
Solve for y as a function of x.
 
A) $y = - ( x + 1 ) ^ { 2 } + C$
B)   $y = - \frac { 2 } { x ^ { 2 } + C }$
C)   $y = - \frac { 2 } { x ( x + 2 ) + C }$
D)   $y = \frac { C } { ( x + 1 ) ^ { 2 } }$
E)   $y = - \frac { 2 } { x + C }$

Quizplus · Accepted Answer

Separating variables and integrating, we get$\int \frac { \mathrm { d } y } { y ^ { 2 } } = \int ( x + 1 ) \mathrm { d } x$$\Rightarrow - \frac { 1 } { y } = \frac { x ^ { 2 } } { 2 } + x + C$$\Rightarrow y = - \frac { 2 } { x ^ { 2 } + 2 x + C }$$\Rightarrow y = - \frac { 2 } { x ( x + 2 ) + C }$

Find the General Solution of the Differential Equation dy dx=(x+1)y2\frac { \mathrm { d } y } { \mathrm {~d} x } = ( x + 1 ) y ^ { 2 } dxdy​=(x+1)y2

Find the General Solution of the Differential Equation $\frac { \mathrm { d } y } { \mathrm {~d} x } = ( x + 1 ) y ^ { 2 }$