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Popular >

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

Solution

d x - \frac{d y}{1 + y ^{2}} = 0

Solution

y = tan (x + c_{1})

Solution steps

d x - \frac{d y}{1 + y ^{2}} = 0

Solve separable ODE:

y = tan (x + c_{1})

y = tan (x + c_{1})

Popular Examples

(dy)/(dx)=(2xy)/(x^2+1),y(1)=2

\frac{d y}{d x} = \frac{2 x y}{x ^{2} + 1}, y (1) = 2

xdx-ydy=xy(xdy-ydx)

x d x - y d y = x y (x d y - y d x)

(dy)/(dx)=(sin(x))/(sec^2(y))

\frac{d y}{d x} = \frac{sin ( x )}{sec ^{2} ( y )}

y^'=(y^4)/(x^6)

y^{^{'}} = \frac{y ^{4}}{x ^{6}}

(dp)/(dt)=p*(1-p*10^{-4})

\frac{d p}{d t} = p \cdot (1 - p \cdot 1 0^{- 4})

Frequently Asked Questions (FAQ)

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