What is the solution for y^'+yt=6cos(5t) ?

The solution for y^'+yt=6cos(5t) is y=(integral of 6cos(5t)e^{(t^2)/2})/(e^{(t^2)/2)}

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y^'+yt=6cos(5t)

y^{^{'}} + y t = 6 cos (5 t)

y = \frac{\int 6 cos ( 5 t ) e ^{\frac{t ^{2}}{2}} d t}{e ^{\frac{t ^{2}}{2}}}

Solution steps

y^{'} (t) + y t = 6 cos (5 t)

Solve linear ODE:

y = \frac{\int 6 cos ( 5 t ) e ^{\frac{t ^{2}}{2}} d t}{e ^{\frac{t ^{2}}{2}}}

y = \frac{\int 6 cos ( 5 t ) e ^{\frac{t ^{2}}{2}} d t}{e ^{\frac{t ^{2}}{2}}}

sin (9 x)

\frac{d}{d x} ((a + \frac{b}{x ^{2}})^{3})

sin (x) ln (x)

f (x) = 2 sin^{5} (x)

x sin (x^{2})

What is the solution for y^'+yt=6cos(5t) ?
The solution for y^'+yt=6cos(5t) is y=(integral of 6cos(5t)e^{(t^2)/2})/(e^{(t^2)/2)}