(2y-1/x+cos(3x))(dy)/(dx)+y/(x^2)-4x^3-3ysin(3x)=0

解

(2 y - \frac{1}{x} + cos (3 x)) \frac{d y}{d x} + \frac{y}{x ^{2}} - 4 x^{3} - 3 y sin (3 x) = 0

y = \frac{- x cos ( 3 x ) + 1 + ( x cos ( 3 x ) - 1 ) ^{2} - 4 x ( - x ^{5} - c _{1} x )}{2 x}, y = \frac{- x cos ( 3 x ) + 1 - ( x cos ( 3 x ) - 1 ) ^{2} - 4 x ( - x ^{5} - c _{1} x )}{2 x}

解答ステップ

(2 y - \frac{1}{x} + cos (3 x)) \frac{d y}{d x} + \frac{y}{x ^{2}} - 4 x^{3} - 3 y sin (3 x) = 0

完全 ODE を解く:

y = \frac{- x cos ( 3 x ) + 1 + ( x cos ( 3 x ) - 1 ) ^{2} - 4 x ( - x ^{5} - c _{1} x )}{2 x}, y = \frac{- x cos ( 3 x ) + 1 - ( x cos ( 3 x ) - 1 ) ^{2} - 4 x ( - x ^{5} - c _{1} x )}{2 x}

y = \frac{- x cos ( 3 x ) + 1 + ( x cos ( 3 x ) - 1 ) ^{2} - 4 x ( - x ^{5} - c _{1} x )}{2 x}, y = \frac{- x cos ( 3 x ) + 1 - ( x cos ( 3 x ) - 1 ) ^{2} - 4 x ( - x ^{5} - c _{1} x )}{2 x}