What is the solution for y^{''}+y=sec^3(x),y(0)=1,y^'(0)= 1/2 ?

The solution for y^{''}+y=sec^3(x),y(0)=1,y^'(0)= 1/2 is y=cos(x)+1/2 sin(x)+(sin(x)tan(x))/2

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y^{''}+y=sec^3(x),y(0)=1,y^'(0)= 1/2

y^{^{''}} + y = sec^{3} (x), y (0) = 1, y^{^{'}} (0) = \frac{1}{2}

y = cos (x) + \frac{1}{2} sin (x) + \frac{sin ( x ) tan ( x )}{2}

Solution steps

y^{''} (x) + y = sec^{3} (x)

Solve linear ODE:

y = cos (x) + \frac{1}{2} sin (x) + \frac{sin ( x ) tan ( x )}{2}

y = cos (x) + \frac{1}{2} sin (x) + \frac{sin ( x ) tan ( x )}{2}

y^{^{''}} - y = x^{2}

y^{^{''}} + y = 2, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} - y = \frac{4 e ^{- x}}{1 - e ^{- 2 x}}

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{1}{3}} e^{x}

y^{^{'''}} + y^{^{'}} = e^{x}

What is the solution for y^{''}+y=sec^3(x),y(0)=1,y^'(0)= 1/2 ?
The solution for y^{''}+y=sec^3(x),y(0)=1,y^'(0)= 1/2 is y=cos(x)+1/2 sin(x)+(sin(x)tan(x))/2