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Popular >

y^'=((t-y))/2

Solution

y^{^{'}} = \frac{( t - y )}{2}

Solution

y = t - 2 + \frac{c _{1}}{e ^{\frac{t}{2}}}

Solution steps

y^{'} (t) = \frac{( t - y )}{2}

Solve linear ODE:

y = t - 2 + \frac{c _{1}}{e ^{\frac{t}{2}}}

y = t - 2 + \frac{c _{1}}{e ^{\frac{t}{2}}}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=((t-y))/2 ?
The solution for y^'=((t-y))/2 is y=t-2+(c_{1})/(e^{t/2)}