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Popular >

(1+y^2)dx-xdy=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)e^{x+y}=x+1

\frac{d y}{d x} e^{x + y} = x + 1

(dy)/(dx)=e^x(1-y)^2

\frac{d y}{d x} = e^{x} (1 - y)^{2}

y^{^{'}} = sin (3 y)

(dy)/(dx)=e^{5y}cos(x/2)

\frac{d y}{d x} = e^{5 y} cos (\frac{x}{2})

(x^2-3)(dy)/(dx)-x(y-1)=0

(x^{2} - 3) \frac{d y}{d x} - x (y - 1) = 0

Frequently Asked Questions (FAQ)

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