What is the solution for y*y^{''}+(y)^2=0,y(0)=4,y^'(0)=6 ?

The solution for y*y^{''}+(y)^2=0,y(0)=4,y^'(0)=6 is y=4cos(t)+6sin(t)

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y*y^{''}+(y)^2=0,y(0)=4,y^'(0)=6

y \cdot y^{^{''}} + (y)^{2} = 0, y (0) = 4, y^{^{'}} (0) = 6

y = 4 cos (t) + 6 sin (t)

Solution steps

y \cdot y^{''} (t) + (y)^{2} = 0

Solve linear ODE:

y = 4 cos (t) + 6 sin (t)

y = 4 cos (t) + 6 sin (t)

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = - 3, y^{^{'}} (0) = 33

y^{^{''}} - y^{^{'}} + 2 y = 0, y (0) = 5

x^{^{''}} + c^{2} \cdot x = 0

y^{^{'''}} - 8 y^{^{''}} - y^{^{'}} + 8 y = 0

a \frac{d ^{2} y}{d x ^{2}} + y b^{2} = 0

What is the solution for y*y^{''}+(y)^2=0,y(0)=4,y^'(0)=6 ?
The solution for y*y^{''}+(y)^2=0,y(0)=4,y^'(0)=6 is y=4cos(t)+6sin(t)