What is the solution for x^{''}-7x^'+12=12,x(0)=5,x^'(0)=5 ?

The solution for x^{''}-7x^'+12=12,x(0)=5,x^'(0)=5 is x=(5-30/7)e^{7t}+30/7

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x^{''}-7x^'+12=12,x(0)=5,x^'(0)=5

x^{^{''}} - 7 x^{^{'}} + 12 = 12, x (0) = 5, x^{^{'}} (0) = 5

x = (5 - \frac{30}{7}) e^{7 t} + \frac{30}{7}

Solution steps

x^{''} (t) - 7 x^{'} (t) + 12 = 12

Solve linear ODE:

x = (5 - \frac{30}{7}) e^{7 t} + \frac{30}{7}

x = (5 - \frac{30}{7}) e^{7 t} + \frac{30}{7}

y^{^{''}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y^{^{'''}} + 12 y^{^{'}} + 36 y^{^{'}} = 0

y^{^{''}} + 0 y^{^{'}} + 9 y = 0

y^{^{''}} + a y = 0, y (0) = 0, y (5) = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for x^{''}-7x^'+12=12,x(0)=5,x^'(0)=5 ?
The solution for x^{''}-7x^'+12=12,x(0)=5,x^'(0)=5 is x=(5-30/7)e^{7t}+30/7