What is the solution for cos(t)y^'-sin(t)y=t^2 ?

The solution for cos(t)y^'-sin(t)y=t^2 is y=(t^3)/(3cos(t))+(c_{1})/(cos(t))

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cos(t)y^'-sin(t)y=t^2

cos (t) y^{^{'}} - sin (t) y = t^{2}

y = \frac{t ^{3}}{3 cos ( t )} + \frac{c _{1}}{cos ( t )}

Solution steps

cos (t) y^{'} (t) - sin (t) y = t^{2}

Solve linear ODE:

y = \frac{t ^{3}}{3 cos ( t )} + \frac{c _{1}}{cos ( t )}

y = \frac{t ^{3}}{3 cos ( t )} + \frac{c _{1}}{cos ( t )}

\frac{x}{x + \frac{c}{x}}

7 x + 6 x

f (x) = 5 x^{\frac{3}{5}} - 4

\int \frac{I n x}{( x + 1 ) ^{2}} d x

x \to \infty lim (4 - x^{2})

What is the solution for cos(t)y^'-sin(t)y=t^2 ?
The solution for cos(t)y^'-sin(t)y=t^2 is y=(t^3)/(3cos(t))+(c_{1})/(cos(t))