What is the solution for y^'=3-2x-0.5y,y(0)=1 ?

The solution for y^'=3-2x-0.5y,y(0)=1 is y=-4x+14-(13)/(e^{0.5x)}

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y^'=3-2x-0.5y,y(0)=1

y^{^{'}} = 3 - 2 x - 0.5 y, y (0) = 1

y = - 4 x + 14 - \frac{13}{e ^{0.5 x}}

Solution steps

y^{'} (x) = 3 - 2 x - 0.5 y

Solve linear ODE:

y = - 4 x + 14 - \frac{13}{e ^{0.5 x}}

y = - 4 x + 14 - \frac{13}{e ^{0.5 x}}

\frac{d}{d x} ((ln (x))^{s e c (x)})

\frac{\partial}{\partial y} (x \cdot y^{2} \cdot e^{z})

\int \frac{8}{( x - 5 ) ( x + 3 )} d x

\int e^{x} \cdot ln (x) d x

n = 1 \sum \infty \frac{n}{e ^{n^{2}}}

What is the solution for y^'=3-2x-0.5y,y(0)=1 ?
The solution for y^'=3-2x-0.5y,y(0)=1 is y=-4x+14-(13)/(e^{0.5x)}