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

The solution for y^{''}-4y^'+3y=x is y=c_{1}e^{3x}+c_{2}e^x+x/3+4/9

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y^{''}-4y^'+3y=x

y^{^{''}} - 4 y^{^{'}} + 3 y = x

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{x}{3} + \frac{4}{9}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = x

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{x}{3} + \frac{4}{9}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{x}{3} + \frac{4}{9}

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{3}}

x \to 0 lim (\frac{arcsin ( x )}{7 x})

f (x) = sin^{4} (x), g (x) = sin^{2} (x)

t \to 0 lim (\frac{e ^{8 t} - 1}{t})

\int e^{x} 2 x d x

What is the solution for y^{''}-4y^'+3y=x ?
The solution for y^{''}-4y^'+3y=x is y=c_{1}e^{3x}+c_{2}e^x+x/3+4/9