What is the solution for -2y^{''}+4y^'+3y=1 ?

The solution for -2y^{''}+4y^'+3y=1 is y=c_{1}e^{-((-2+sqrt(10))t)/2}+c_{2}e^{((2+sqrt(10))t)/2}+1/3

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-2y^{''}+4y^'+3y=1

- 2 y^{^{''}} + 4 y^{^{'}} + 3 y = 1

y = c_{1} e^{- \frac{( - 2 + 10 ) t}{2}} + c_{2} e^{\frac{( 2 + 10 ) t}{2}} + \frac{1}{3}

Solution steps

- 2 y^{''} (t) + 4 y^{'} (t) + 3 y = 1

Solve linear ODE:

y = c_{1} e^{- \frac{( - 2 + 10 ) t}{2}} + c_{2} e^{\frac{( 2 + 10 ) t}{2}} + \frac{1}{3}

y = c_{1} e^{- \frac{( - 2 + 10 ) t}{2}} + c_{2} e^{\frac{( 2 + 10 ) t}{2}} + \frac{1}{3}

f (x) = (x^{2} + 2) (x^{4} + 1)

cos (t) y^{^{'}} - sin (t) y = t^{2}

\frac{x}{x + \frac{c}{x}}

7 x + 6 x

f (x) = 5 x^{\frac{3}{5}} - 4

What is the solution for -2y^{''}+4y^'+3y=1 ?
The solution for -2y^{''}+4y^'+3y=1 is y=c_{1}e^{-((-2+sqrt(10))t)/2}+c_{2}e^{((2+sqrt(10))t)/2}+1/3