What is the solution for y^'+2/3 y=1-1/2 t ?

The solution for y^'+2/3 y=1-1/2 t is y= 3/2-(3(2t-3))/8+(c_{1}}{e^{\frac{2t)/3}}

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y^'+2/3 y=1-1/2 t

y^{^{'}} + \frac{2}{3} y = 1 - \frac{1}{2} t

y = \frac{3}{2} - \frac{3 ( 2 t - 3 )}{8} + \frac{c _{1}}{e ^{\frac{2 t}{3}}}

Solution steps

y^{'} (t) + \frac{2}{3} y = 1 - \frac{1}{2} t

Solve linear ODE:

y = \frac{3}{2} - \frac{3 ( 2 t - 3 )}{8} + \frac{c _{1}}{e ^{\frac{2 t}{3}}}

y = \frac{3}{2} - \frac{3 ( 2 t - 3 )}{8} + \frac{c _{1}}{e ^{\frac{2 t}{3}}}

\frac{d}{d x} (A (4 e^{2 x} x + 4 e^{2 x}))

\frac{d}{d x} ((x^{5} + 2 x) (x^{3} + x^{2} + x + 7))

\int \frac{1}{( x + 3 ) ^{2}} d x

y = 9 x^{2} ln (x), y = 36 ln (x)

cos^{2} (t)

What is the solution for y^'+2/3 y=1-1/2 t ?
The solution for y^'+2/3 y=1-1/2 t is y= 3/2-(3(2t-3))/8+(c_{1}}{e^{\frac{2t)/3}}