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Popular >

yy^'=t+ty

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

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Frequently Asked Questions (FAQ)

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