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Popular >

ty^'+y=t^3,y(1)=-2

Solution

t y^{^{'}} + y = t^{3}, y (1) = - 2

Solution

y = \frac{t ^{3}}{4} - \frac{9}{4 t}

Solution steps

t y^{'} (t) + y = t^{3}

Solve linear ODE:

y = \frac{t ^{3}}{4} - \frac{9}{4 t}

y = \frac{t ^{3}}{4} - \frac{9}{4 t}

Popular Examples

integral from 1/2 to 1 of (x^{-3}-3)

\int_{\frac{1}{2}}^{1} (x^{- 3} - 3) d x

limit as x approaches 2 of x^3+2x^2+1

x \to 2 lim (x^{3} + 2 x^{2} + 1)

implicit (dy)/(dx),x^3-3xy^2+y^3=1 implicit derivative

\frac{d y}{d x}, x^{3} - 3 x y^{2} + y^{3} = 1

y^'=3x+y,y(0)=5

y^{^{'}} = 3 x + y, y (0) = 5

integral from 0 to 1 of x/(x^2+4x+13)

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 13} d x

Frequently Asked Questions (FAQ)

What is the solution for ty^'+y=t^3,y(1)=-2 ?
The solution for ty^'+y=t^3,y(1)=-2 is y=(t^3)/4-9/(4t)