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Popular >

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

Solution

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

Solution

y = \frac{1}{x} - \frac{e ^{1 - x}}{x}

Solution steps

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

Solve linear ODE:

y = \frac{1}{x} - \frac{e ^{1 - x}}{x}

y = \frac{1}{x} - \frac{e ^{1 - x}}{x}

Graph

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Frequently Asked Questions (FAQ)

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