What is the solution for y^'=(5x+1)(y+11),y(0)=0 ?

The solution for y^'=(5x+1)(y+11),y(0)=0 is y=11e^{(5x^2)/2+x}-11

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y^'=(5x+1)(y+11),y(0)=0

y^{^{'}} = (5 x + 1) (y + 11), y (0) = 0

y = 11 e^{\frac{5 x ^{2}}{2} + x} - 11

Solution steps

y^{'} (x) = (5 x + 1) (y + 11)

Solve separable ODE:

y = 11 e^{\frac{5 x ^{2}}{2} + x} - 11

y = 11 e^{\frac{5 x ^{2}}{2} + x} - 11

x^{2} sin (x d x) + x y d y = 0

e^{2 x} (2 cos (y) d x - sin (y) d y) = 0, y (0) = 0

sinh (x) d x + y^{- 1} cosh (x) d y = 0

\frac{d y}{d x} = \frac{2 cos ( 2 x )}{3 + 2 y}

\frac{d y}{d x} = \frac{- x}{y - 3}

What is the solution for y^'=(5x+1)(y+11),y(0)=0 ?
The solution for y^'=(5x+1)(y+11),y(0)=0 is y=11e^{(5x^2)/2+x}-11