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Popular >

xy^'+y+x=0,y(1)=1

Solution

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

Solution

y = - \frac{x}{2} + \frac{3}{2 x}

Solution steps

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

Solve linear ODE:

y = - \frac{x}{2} + \frac{3}{2 x}

y = - \frac{x}{2} + \frac{3}{2 x}

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\int \frac{1}{4} x^{- \frac{3}{4}} d x

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Frequently Asked Questions (FAQ)

What is the solution for xy^'+y+x=0,y(1)=1 ?
The solution for xy^'+y+x=0,y(1)=1 is y=-x/2+3/(2x)