次の微分方程式の一般解を求めなさい.
dy dx =2
dy dx =x
dy dx =3y
dy dx =- 2y
dy dx =− 2x y
dy dx = 1+y 1+x
dy dx =2y+3
dy dx =3 x 2 y
dy dx =4 x 2
dy dx =3x+1
dy dx =5 x 3 −7 x 2 −x+5
dy dx =xy
dy dx = y x
dy dx =sinx
dy dx =cosx
dy dx =4sin5x
dy dx =6cos2x
dy dx =2 e 3x
dy dx =− cosx siny
dy dx = 1+ e x e x−y
( tan −1 y ) y ′ = 2− x 2 x 2 −2
y 2 dx− x 3 dy=0
x−1 y ′ = 1 y
y ′ = x y 3 +x y 2 x 3 y−xy
y ′ = sin 2 xsinycosx
dy dx = e y e x
( y 2 +siny ) y ′ +cosx+ x 3 =0
次の微分方程式を( )内の初期条件で解け
dy dx = 1 2xy
( x=1 , y=1 )
dy dx = e 2x e 3y ( x=0 , y=0 )
dy dx =sinx cos 2 y ( x=0 , y= π 4 )