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Popular >

y^'=y+t

Solution

y^{^{'}} = y + t

Solution

y = - t - 1 + c_{1} e^{t}

Solution steps

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

Solve linear ODE:

y = - t - 1 + c_{1} e^{t}

y = - t - 1 + c_{1} e^{t}

Popular Examples

integral from 0 to pi of sin^2(xco)s^2x

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

tangent of y=sqrt(x^2-x+5),\at x=5 tangent of

y = x^{2} - x + 5, at x = 5

integral of (2x+1)/(x^2-2x+5)

\int \frac{2 x + 1}{x ^{2} - 2 x + 5} d x

integral of csc^4(θ)

\int csc^{4} (θ) d θ

derivative of |x+3|

\frac{d}{d x} (∣ x + 3 ∣)

Frequently Asked Questions (FAQ)

What is the solution for y^'=y+t ?
The solution for y^'=y+t is y=-t-1+c_{1}e^t