The solution of the differential equation $x \frac { d y } { d x } + 2 y = x ^ { 2 } , ( x \neq 0 )$ with $y ( 1 ) = 1$, is
(1) $y = \frac { x ^ { 3 } } { 5 } + \frac { 1 } { 5 x ^ { 2 } }$
(2) $y = \frac { 3 } { 4 } x ^ { 2 } + \frac { 1 } { 4 x ^ { 2 } }$
(3) $y = \frac { x ^ { 2 } } { 4 } + \frac { 3 } { 4 x ^ { 2 } }$
(4) $y = \frac { 4 } { 5 } x ^ { 3 } + \frac { 1 } { 5 x ^ { 2 } }$

If $\cos x \frac { d y } { d x } - y \sin x = 6 x , \left( 0 < x < \frac { \pi } { 2 } \right)$ and $y \left( \frac { \pi } { 3 } \right) = 0$, then $y \left( \frac { \pi } { 6 } \right)$ is equal to
(1) $- \frac { \pi ^ { 2 } } { 4 \sqrt { 3 } }$
(2) $\frac { \pi ^ { 2 } } { 2 \sqrt { 3 } }$
(3) $- \frac { \pi ^ { 2 } } { 2 }$
(4) $- \frac { \pi ^ { 2 } } { 2 \sqrt { 3 } }$

If the solution curve of the differential equation $\left( 2 x - 10 y ^ { 3 } \right) d y + y d x = 0$, passes through the points $( 0,1 )$ and $( 2 , \beta )$, then $\beta$ is a root of the equation? (1) $y ^ { 5 } - 2 y - 2 = 0$ (2) $y ^ { 5 } - y ^ { 2 } - 1 = 0$ (3) $2 y ^ { 5 } - y ^ { 2 } - 2 = 0$ (4) $2 y ^ { 5 } - 2 y - 1 = 0$

If $y = y ( x )$ is the solution of the differential equation $x \frac { d y } { d x } + 2 y = x e ^ { x } , y ( 1 ) = 0$ then the local maximum value of the function $z ( x ) = x ^ { 2 } y ( x ) - e ^ { x } , x \in R$ is
(1) $1 - e$
(2) 0
(3) $\frac { 1 } { 2 }$
(4) $\frac { 4 } { e } - e$

If $\frac { d y } { d x } + e ^ { x } \left( x ^ { 2 } - 2 \right) y = \left( x ^ { 2 } - 2 x \right) \left( x ^ { 2 } - 2 \right) e ^ { 2 x }$ and $y ( 0 ) = 0$, then the value of $y ( 2 )$ is
(1) $-1$
(2) 1
(3) 0
(4) $e$

Let the solution curve $y = y ( x )$ of the differential equation $\left( 4 + x ^ { 2 } \right) dy - 2 x \left( x ^ { 2 } + 3 y + 4 \right) dx = 0$ pass through the origin. Then $y ( 2 )$ is equal to $\_\_\_\_$.

The general solution of the differential equation $(x - y^2)dx + y(5x + y^2)dy = 0$ is
(1) $y ^ { 2 } + x ^ { 4 } = C(y ^ { 2 } + 2x ^ { 3 })$
(2) $y ^ { 2 } + 2x ^ { 4 } = C(y ^ { 2 } + x ^ { 3 })$
(3) $y ^ { 2 } + x ^ { 3 } = C(2y ^ { 2 } + x ^ { 4 })$
(4) $y ^ { 2 } + 2x ^ { 3 } = C(2y ^ { 2 } + x ^ { 4 })$

Let $y = y ( x )$ be the solution curve of the differential equation $\frac { d y } { d x } + \frac { 1 } { x ^ { 2 } - 1 } y = \left( \frac { x - 1 } { x + 1 } \right) ^ { \frac { 1 } { 2 } } , x > 1$ passing through the point $\left( 2 , \sqrt { \frac { 1 } { 3 } } \right)$. Then $\sqrt { 7 } y ( 8 )$ is equal to
(1) $11 + 6 \log _ { e } 3$
(2) $19$
(3) $12 - 2 \log _ { e } 3$
(4) $19 - 6 \log _ { e } 3$

A function $y = f ( x )$ satisfies $f ( x ) \sin 2 x + \sin x - \left( 1 + \cos ^ { 2 } x \right) f ^ { \prime } ( x ) = 0$ with condition $f ( 0 ) = 0$. Then $f \left( \frac { \pi } { 2 } \right)$ is equal to
(1) 1
(2) 0
(3) - 1
(4) 2

Let $y = y ( x )$ be the solution of the differential equation $\left( x ^ { 2 } + 4 \right) ^ { 2 } d y + \left( 2 x ^ { 3 } y + 8 x y - 2 \right) d x = 0$. If $y ( 0 ) = 0$, then $y ( 2 )$ is equal to
(1) $\frac { \pi } { 32 }$
(2) $2 \pi$
(3) $\frac { \pi } { 8 }$
(4) $\frac { \pi } { 16 }$

Let $y = f ( x )$ be the solution of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { x y } { x ^ { 2 } - 1 } = \frac { x ^ { 6 } + 4 x } { \sqrt { 1 - x ^ { 2 } } } , - 1 < x < 1$ such that $f ( 0 ) = 0$. If $6 \int _ { - 1 / 2 } ^ { 1 / 2 } f ( x ) \mathrm { d } x = 2 \pi - \alpha$ then $\alpha ^ { 2 }$ is equal to $\_\_\_\_$