What is the solution for y^'+2y=4cos(2x),y(1/4 pi)=3 ?

The solution for y^'+2y=4cos(2x),y(1/4 pi)=3 is y=sin(2x)+cos(2x)+2e^{(pi-4x)/2}

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y^'+2y=4cos(2x),y(1/4 pi)=3

y^{^{'}} + 2 y = 4 cos (2 x), y (\frac{1}{4} π) = 3

y = sin (2 x) + cos (2 x) + 2 e^{\frac{π - 4 x}{2}}

Solution steps

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

Solve linear ODE:

y = sin (2 x) + cos (2 x) + 2 e^{\frac{π - 4 x}{2}}

y = sin (2 x) + cos (2 x) + 2 e^{\frac{π - 4 x}{2}}

n = 6 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

\int_{0}^{2} 3 x + 5 d x

\int \frac{1}{x ^{4} + 4} d x

y = \frac{x ^{2} - 2 x}{x}

\frac{\partial}{\partial v} (2 v)

What is the solution for y^'+2y=4cos(2x),y(1/4 pi)=3 ?
The solution for y^'+2y=4cos(2x),y(1/4 pi)=3 is y=sin(2x)+cos(2x)+2e^{(pi-4x)/2}