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

The solution for y^'+y= 1/((cos(2x))^{3/2)} is y=(integral of sec^{3/2}(2x)e^x)/(e^x)

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y^'+y= 1/((cos(2x))^{3/2)}

y^{^{'}} + y = \frac{1}{( cos ( 2 x ) ) ^{\frac{3}{2}}}

y = \frac{\int sec ^{\frac{3}{2}} ( 2 x ) e ^{x} d x}{e ^{x}}

Solution steps

y^{'} (x) + y = \frac{1}{( cos ( 2 x ) ) ^{\frac{3}{2}}}

Solve linear ODE:

y = \frac{\int sec ^{\frac{3}{2}} ( 2 x ) e ^{x} d x}{e ^{x}}

y = \frac{\int sec ^{\frac{3}{2}} ( 2 x ) e ^{x} d x}{e ^{x}}

\frac{14}{2 x + 16}

\int \frac{x}{7 - x ^{2}} d x

f (x) = e^{x} (- 4 + 7 x + 4 x^{2}), at x = 0

x \to a lim (ln (x - a))

y = \frac{6 ( x ^{4} - 3 )}{x ^{3}}

What is the solution for y^'+y= 1/((cos(2x))^{3/2)} ?
The solution for y^'+y= 1/((cos(2x))^{3/2)} is y=(integral of sec^{3/2}(2x)e^x)/(e^x)