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Popular >

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

Solution

t y^{^{'}} + t y = 1 - y, y (1) = 0

Solution

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

Solution steps

t y^{'} (t) + t y = 1 - y

Solve linear ODE:

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

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

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\frac{d}{d x} (\frac{2 - 3 x}{7 - x})

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\int (e^{x} cos (y) - x - 2) d y

Frequently Asked Questions (FAQ)

What is the solution for ty^'+ty=1-y,y(1)=0 ?
The solution for ty^'+ty=1-y,y(1)=0 is y= 1/t-(e^{1-t})/t