What is the solution for y^{''}=a(y^')^2+b ?

The solution for y^{''}=a(y^')^2+b is y=-(ln(cos(sqrt(a)sqrt(b)t+c_{1}sqrt(a)sqrt(b))))/a+c_{2}

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y^{''}=a(y^')^2+b

y^{^{''}} = a (y^{^{'}})^{2} + b

y = - \frac{ln ( cos ( a b t + c _{1} a b ) )}{a} + c_{2}

Solution steps

y^{''} (t) = a (y^{'} (t))^{2} + b

Solve using substitution:

y = - \frac{ln ( cos ( a b t + c _{1} a b ) )}{a} + c_{2}

y = - \frac{ln ( cos ( a b t + c _{1} a b ) )}{a} + c_{2}

4 e^{3 x} cos (4 x)

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What is the solution for y^{''}=a(y^')^2+b ?
The solution for y^{''}=a(y^')^2+b is y=-(ln(cos(sqrt(a)sqrt(b)t+c_{1}sqrt(a)sqrt(b))))/a+c_{2}