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Popular Calculus >

y^'+2/x y=x,y(1)=0

Solution

y^{^{'}} + \frac{2}{x} y = x, y (1) = 0

Solution

y = \frac{x ^{2}}{4} - \frac{1}{4 x ^{2}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{4} - \frac{1}{4 x ^{2}}

y = \frac{x ^{2}}{4} - \frac{1}{4 x ^{2}}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+2/x y=x,y(1)=0 ?
The solution for y^'+2/x y=x,y(1)=0 is y=(x^2)/4-1/(4x^2)