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Popular >

y^'=1+y

Solution

y^{^{'}} = 1 + y

Solution

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

Solution steps

y^{'} (t) = 1 + y

Solve separable ODE:

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

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

Popular Examples

integral from 0 to 1 of 1/(x-1)

\int_{0}^{1} \frac{1}{x - 1} d x

y^'=((xy+3x-y-3))/((xy-2x+4y-8))

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

inverse oflaplace 4/((3s-1)^5)inverse laplace

\frac{4}{( 3 s - 1 ) ^{5}}

integral of 1/(5+4sec(x))

\int \frac{1}{5 + 4 sec ( x )} d x

y^{^{'}} + \frac{y}{x} = 2 y^{2}

Frequently Asked Questions (FAQ)

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