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

The solution for y^'=(xysin(x))/(y+1),y(0)=1 is y+ln(y)=-xcos(x)+sin(x)+1

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y^'=(xysin(x))/(y+1),y(0)=1

y^{^{'}} = \frac{x y sin ( x )}{y + 1}, y (0) = 1

y + ln (y) = - x cos (x) + sin (x) + 1

Solution steps

y^{'} (x) = \frac{x y sin ( x )}{y + 1}

Solve separable ODE:

y + ln (y) = - x cos (x) + sin (x) + 1

y + ln (y) = - x cos (x) + sin (x) + 1

x^{^{'}} = 1 - 3 x + x^{2}

u^{^{'}} = 3 e^{8 x - H (x)}

y^{^{'}} = \frac{( 3 x ^{2} + 4 x - 4 )}{2 y - 4}, y (1) = 3

(\frac{y}{sec ^{2} ( x )}) \frac{d y}{d x} = 2 tan (x) + 1

x y d y = (1 + y^{2}) d x, y (2) = 3

What is the solution for y^'=(xysin(x))/(y+1),y(0)=1 ?
The solution for y^'=(xysin(x))/(y+1),y(0)=1 is y+ln(y)=-xcos(x)+sin(x)+1