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Popular >

y^'(1+x^2)+(1+y^2)=0

Solution

y^{^{'}} (1 + x^{2}) + (1 + y^{2}) = 0

Solution

y = tan (- x - c_{1})

Solution steps

y^{'} (x) + (1 + y^{2}) = 0

Solve separable ODE:

y = tan (- x - c_{1})

y = tan (- x - c_{1})

Graph

Popular Examples

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x \to - 2 + lim (\frac{2}{x ^{2} - 4})

x(dy)/(dx)+y=x^4y^3

x \frac{d y}{d x} + y = x^{4} y^{3}

y^{^{'}} + 2 x^{2} = 0

integral of (3+x+x^2)/(sqrt(x))

\int \frac{3 + x + x ^{2}}{x} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'(1+x^2)+(1+y^2)=0 ?
The solution for y^'(1+x^2)+(1+y^2)=0 is y=tan(-x-c_{1})