What is the solution for y^'y=x-3-6cos(x) ?

The solution for y^'y=x-3-6cos(x) is y=sqrt(x^2-6x-12sin(x)+c_{1)},y=-sqrt(x^2-6x-12sin(x)+c_{1)}

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y^'y=x-3-6cos(x)

y^{^{'}} y = x - 3 - 6 cos (x)

y = x^{2} - 6 x - 12 sin (x) + c_{1}, y = - x^{2} - 6 x - 12 sin (x) + c_{1}

Solution steps

y^{'} (x) y = x - 3 - 6 cos (x)

Solve separable ODE:

y = x^{2} - 6 x - 12 sin (x) + c_{1}, y = - x^{2} - 6 x - 12 sin (x) + c_{1}

y = x^{2} - 6 x - 12 sin (x) + c_{1}, y = - x^{2} - 6 x - 12 sin (x) + c_{1}

What is the solution for y^'y=x-3-6cos(x) ?
The solution for y^'y=x-3-6cos(x) is y=sqrt(x^2-6x-12sin(x)+c_{1)},y=-sqrt(x^2-6x-12sin(x)+c_{1)}