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Popular >

y^'=(1+x)(1+y)

Solution

y^{^{'}} = (1 + x) (1 + y)

Solution

y = e^{x + \frac{x ^{2}}{2} + c_{1}} - 1

Solution steps

y^{'} (x) = (1 + x) (1 + y)

Solve separable ODE:

y = e^{x + \frac{x ^{2}}{2} + c_{1}} - 1

y = e^{x + \frac{x ^{2}}{2} + c_{1}} - 1

Popular Examples

limit as x approaches 0 of ((5e^x-5))/x

x \to 0 lim (\frac{( 5 e ^{x} - 5 )}{x})

(\partial)/(\partial y)(x^2-3y^2+7)

\frac{\partial}{\partial y} (x^{2} - 3 y^{2} + 7)

integral of ((x+1))/((x-1))

\int \frac{( x + 1 )}{( x - 1 )} d x

limit as x approaches 1+of 2

x \to 1 + lim (2)

tangent of f(x)=sqrt(x),(16,4)tangent of

f (x) = x, (16, 4)

Frequently Asked Questions (FAQ)

What is the solution for y^'=(1+x)(1+y) ?
The solution for y^'=(1+x)(1+y) is y=e^{x+(x^2)/2+c_{1}}-1