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Popular >

y^'=t+3

Solution

y^{^{'}} = t + 3

Solution

y = \frac{t ^{2}}{2} + 3 t + c_{1}

Solution steps

y^{'} (t) = t + 3

If

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

then

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

y = \int t + 3 d t

\int t + 3 d t = \frac{t ^{2}}{2} + 3 t + c_{1}

y = \frac{t ^{2}}{2} + 3 t + c_{1}

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tangent of y=cos(x),\at x= pi/4 tangent of

y = cos (x), at x = \frac{π}{4}

y^{^{''}} + y^{^{'}} - y = 0

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Frequently Asked Questions (FAQ)

What is the solution for y^'=t+3 ?
The solution for y^'=t+3 is y=(t^2)/2+3t+c_{1}