What is the solution for y^{''}-4y^'+3y=x,y(0)=6,y^'(0)=10 ?

The solution for y^{''}-4y^'+3y=x,y(0)=6,y^'(0)=10 is y=(-7/2+50/9)e^{3x}+7/2 e^x+x/3+4/9

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y^{''}-4y^'+3y=x,y(0)=6,y^'(0)=10

y^{^{''}} - 4 y^{^{'}} + 3 y = x, y (0) = 6, y^{^{'}} (0) = 10

y = (- \frac{7}{2} + \frac{50}{9}) e^{3 x} + \frac{7}{2} e^{x} + \frac{x}{3} + \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = (- \frac{7}{2} + \frac{50}{9}) e^{3 x} + \frac{7}{2} e^{x} + \frac{x}{3} + \frac{4}{9}

y = (- \frac{7}{2} + \frac{50}{9}) e^{3 x} + \frac{7}{2} e^{x} + \frac{x}{3} + \frac{4}{9}

y^{^{''}} + 6 y^{^{'}} + 25 y = x^{2}

y^{^{''}} + 2 y^{^{'}} - 8 y = - 650 sin (3 t)

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

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 4 x = 3 sin (2 t)

y^{^{''}} + 4 y = cos (x), y (π) = 0, y (0) = 0

What is the solution for y^{''}-4y^'+3y=x,y(0)=6,y^'(0)=10 ?
The solution for y^{''}-4y^'+3y=x,y(0)=6,y^'(0)=10 is y=(-7/2+50/9)e^{3x}+7/2 e^x+x/3+4/9