What is the solution for (2t-1)y^'-4y=0 ?

The solution for (2t-1)y^'-4y=0 is y=4e^{c_{1}}t^2-4e^{c_{1}}t+e^{c_{1}}

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(2t-1)y^'-4y=0

(2 t - 1) y^{^{'}} - 4 y = 0

y = 4 e^{c_{1}} t^{2} - 4 e^{c_{1}} t + e^{c_{1}}

Solution steps

(2 t - 1) y^{'} (t) - 4 y = 0

Solve separable ODE:

y = 4 e^{c_{1}} t^{2} - 4 e^{c_{1}} t + e^{c_{1}}

y = 4 e^{c_{1}} t^{2} - 4 e^{c_{1}} t + e^{c_{1}}

\frac{d p}{d t} + 2 tp = p + 8 t - 4

f (x) f^{^{'}} (x) = x x^{3} - x x^{2} - 6 xx

f^{^{'}} (x) = f (x)^{- 1} (x)

x y^{2} y^{^{'}} = x + x^{3}

\frac{3 d y}{d t} = \frac{cos ( t )}{y}

What is the solution for (2t-1)y^'-4y=0 ?
The solution for (2t-1)y^'-4y=0 is y=4e^{c_{1}}t^2-4e^{c_{1}}t+e^{c_{1}}