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Popular >

y^'=2y-t^2

Solution

y^{^{'}} = 2 y - t^{2}

Solution

y = \frac{t ^{2}}{2} + \frac{t}{2} + \frac{1}{4} + c_{1} e^{2 t}

Solution steps

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

Solve linear ODE:

y = \frac{t ^{2}}{2} + \frac{t}{2} + \frac{1}{4} + c_{1} e^{2 t}

y = \frac{t ^{2}}{2} + \frac{t}{2} + \frac{1}{4} + c_{1} e^{2 t}

Popular Examples

integral from 0 to 2 of sin(x^2)

\int_{0}^{2} sin (x^{2}) d x

(\partial)/(\partial x)(-xyz^2+16)

\frac{\partial}{\partial x} (- x y z^{2} + 16)

tangent of y=x-x^3(1)tangent of

y = x - x^{3} (1)

inverse oflaplace 5/(s^2+4)inverse laplace

\frac{5}{s ^{2} + 4}

(\partial)/(\partial x)(-4x^5y^{-5})

\frac{\partial}{\partial x} (- 4 x^{5} y^{- 5})

Frequently Asked Questions (FAQ)

What is the solution for y^'=2y-t^2 ?
The solution for y^'=2y-t^2 is y=(t^2)/2+t/2+1/4+c_{1}e^{2t}