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Popular >

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

Solution

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

Solution

y = \frac{1}{2} (- x + ln (x)) + c_{1}

Solution steps

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

Isolate

y^{'} (x) : y^{'} (x) = \frac{- x + 1}{2 x}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

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

\int \frac{- x + 1}{2 x} d x = \frac{1}{2} (- x + ln (x)) + c_{1}

y = \frac{1}{2} (- x + ln (x)) + c_{1}

Popular Examples

y^{^{''}} + 2 y^{^{'}} = 3

derivative of y-3x

\frac{d}{d x} (y - 3 x)

derivative of 2x^3+1/(x^2+16x^{7/2}+5)

\frac{d}{d x} (2 x^{3} + \frac{1}{x ^{2}} + 16 x^{\frac{7}{2}} + 5)

2xy^'+y=6x,y(4)=20

2 x y^{^{'}} + y = 6 x, y (4) = 20

limit as x approaches 6-of (x+9)/(x+6)

x \to 6 - lim (\frac{x + 9}{x + 6})

Frequently Asked Questions (FAQ)

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