What is the solution for y^{''}-2/3 y^'+y/9 =0,y(0)=17,y^'(0)=b ?

The solution for y^{''}-2/3 y^'+y/9 =0,y(0)=17,y^'(0)=b is y=17e^{t/3}+(b-17/3)e^{t/3}t

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y^{''}-2/3 y^'+y/9 =0,y(0)=17,y^'(0)=b

y^{^{''}} - \frac{2}{3} y^{^{'}} + \frac{y}{9} = 0, y (0) = 17, y^{^{'}} (0) = b

y = 17 e^{\frac{t}{3}} + (b - \frac{17}{3}) e^{\frac{t}{3}} t

Solution steps

y^{''} (t) - \frac{2}{3} y^{'} (t) + \frac{y}{9} = 0

Solve linear ODE:

y = 17 e^{\frac{t}{3}} + (b - \frac{17}{3}) e^{\frac{t}{3}} t

y = 17 e^{\frac{t}{3}} + (b - \frac{17}{3}) e^{\frac{t}{3}} t

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + 10 y + 16 y = 0

8 \frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 5 y = 0

\frac{d ^{2} y}{d t ^{2}} - 18 \frac{d y}{d t} + 81 y = 0

y^{^{'''}} + 4 y^{^{''}} - 25 y^{^{'}} - 28 y = 0

What is the solution for y^{''}-2/3 y^'+y/9 =0,y(0)=17,y^'(0)=b ?
The solution for y^{''}-2/3 y^'+y/9 =0,y(0)=17,y^'(0)=b is y=17e^{t/3}+(b-17/3)e^{t/3}t