What is the solution for y+2t^2+y^'tln(t)=0 ?

The solution for y+2t^2+y^'tln(t)=0 is y=-(t^2)/(ln(t))+(c_{1})/(ln(t))

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y+2t^2+y^'tln(t)=0

y + 2 t^{2} + y^{^{'}} t ln (t) = 0

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

Solution steps

y + 2 t^{2} + y^{'} (t) t ln (t) = 0

Solve linear ODE:

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

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

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\frac{\partial}{\partial x} (ln (\frac{3 x y}{5 z}))

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∣ 4 x - 1 ∣, x^{2} - 5

What is the solution for y+2t^2+y^'tln(t)=0 ?
The solution for y+2t^2+y^'tln(t)=0 is y=-(t^2)/(ln(t))+(c_{1})/(ln(t))