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Popular >

xu'=u^2+1

Solution

x u^{^{'}} = u^{2} + 1

Solution

u (x) = tan (ln (x) + c_{1})

Solution steps

x u^{'} (x) = u (x)^{2} + 1

Solve separable ODE:

u (x) = tan (ln (x) + c_{1})

u (x) = tan (ln (x) + c_{1})

Popular Examples

y^{^{'}} + 2 y = - 12

(dy)/(dt)=k(650-y),y(0)=300

\frac{d y}{d t} = k (650 - y), y (0) = 300

(dy)/(dx)=(20xy)/((ln(y))^6)

\frac{d y}{d x} = \frac{20 x y}{( ln ( y ) ) ^{6}}

f^'(x)=f(x)(x)^{-1}(x)

f^{^{'}} (x) = f (x) (x)^{- 1} (x)

(dy}{dx}=(\frac{y+1)/x)

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

Frequently Asked Questions (FAQ)

What is the solution for xu'=u^2+1 ?
The solution for xu'=u^2+1 is u(x)=tan(ln(x)+c_{1})