What is the solution for y^'=(3100)/(1+0.25t) ?

The solution for y^'=(3100)/(1+0.25t) is y=12400ln(1+0.25t)+c_{1}

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y^'=(3100)/(1+0.25t)

y^{^{'}} = \frac{3100}{1 + 0.25 t}

y = 12400 ln (1 + 0.25 t) + c_{1}

Solution steps

y^{'} (t) = \frac{3100}{1 + 0.25 t}

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

then

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

y = \int \frac{3100}{1 + 0.25 t} d t

\int \frac{3100}{1 + 0.25 t} d t = 12400 ln (1 + 0.25 t) + c_{1}

y = 12400 ln (1 + 0.25 t) + c_{1}

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What is the solution for y^'=(3100)/(1+0.25t) ?
The solution for y^'=(3100)/(1+0.25t) is y=12400ln(1+0.25t)+c_{1}