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Popular >

4ty^'+4=5+t

Solution

4 t y^{^{'}} + 4 = 5 + t

Solution

y = \frac{1}{4} (t + ln (t)) + c_{1}

Solution steps

4 t y^{'} (t) + 4 = 5 + t

Isolate

y^{'} (t) : y^{'} (t) = \frac{t + 1}{4 t}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{t + 1}{4 t} d t

\int \frac{t + 1}{4 t} d t = \frac{1}{4} (t + ln (t)) + c_{1}

y = \frac{1}{4} (t + ln (t)) + c_{1}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for 4ty^'+4=5+t ?
The solution for 4ty^'+4=5+t is y= 1/4 (t+ln(t))+c_{1}