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Popular Calculus >

x^2y^'+xy=1,y(1)=2

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Graph

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Frequently Asked Questions (FAQ)

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