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Popular >

ty^'+ty=1-y

Solution

t \cdot y^{^{'}} + t \cdot y = 1 - y

Solution

y = \frac{1}{t} + \frac{c _{1}}{e ^{t} t}

Solution steps

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

Solve linear ODE:

y = \frac{1}{t} + \frac{c _{1}}{e ^{t} t}

y = \frac{1}{t} + \frac{c _{1}}{e ^{t} t}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for t*y^'+t*y=1-y ?
The solution for t*y^'+t*y=1-y is y= 1/t+(c_{1})/(e^tt)