What is the solution for y^'=sec^2(x)(2+y)^2,y(pi)=-5 ?

The solution for y^'=sec^2(x)(2+y)^2,y(pi)=-5 is y=-5/(3tan(x)+1)-(6tan(x))/(3tan(x)+1)

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y^'=sec^2(x)(2+y)^2,y(pi)=-5

y^{^{'}} = sec^{2} (x) (2 + y)^{2}, y (π) = - 5

y = - \frac{5}{3 tan ( x ) + 1} - \frac{6 tan ( x )}{3 tan ( x ) + 1}

Solution steps

y^{'} (x) = sec^{2} (x) (2 + y)^{2}

Solve separable ODE:

y = - \frac{5}{3 tan ( x ) + 1} - \frac{6 tan ( x )}{3 tan ( x ) + 1}

y = - \frac{5}{3 tan ( x ) + 1} - \frac{6 tan ( x )}{3 tan ( x ) + 1}

\frac{d}{d x} (8 4 x)

y^{^{''}} + 4 y = 3 sin (2 x)

d er i v a t i v e \frac{9}{x + 2}

\frac{\partial}{\partial x} (x y e^{z} + x - z^{2} - 3)

\int sin^{3} (10 x) d x

What is the solution for y^'=sec^2(x)(2+y)^2,y(pi)=-5 ?
The solution for y^'=sec^2(x)(2+y)^2,y(pi)=-5 is y=-5/(3tan(x)+1)-(6tan(x))/(3tan(x)+1)