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Popular >

y^'=t-2y

Solution

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

Solution

y = \frac{2 t - 1}{4} + \frac{c _{1}}{e ^{2 t}}

Solution steps

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

Solve linear ODE:

y = \frac{2 t - 1}{4} + \frac{c _{1}}{e ^{2 t}}

y = \frac{2 t - 1}{4} + \frac{c _{1}}{e ^{2 t}}

Popular Examples

integral from 2 to infinity of 1/(x^4)

\int_{2}^{\infty} \frac{1}{x ^{4}} d x

tangent of f(x)=x^2-7x+12,\at x=2 tangent of

f (x) = x^{2} - 7 x + 12, at x = 2

derivative of y=sqrt(4-\sqrt{4+x)}derivative of

y = 4 - 4 + x

d/(dr(t))((s-r(t))/t)

\frac{d}{d r ( t )} (\frac{s - r ( t )}{t})

(\partial)/(\partial x)(e-4tcos(pix))

\frac{\partial}{\partial x} (e - 4 t cos (π x))

Frequently Asked Questions (FAQ)

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