What is the solution for 2.25y^{''}-3y^'+y=0,y(0)=2,y^'(0)=-1 ?

The solution for 2.25y^{''}-3y^'+y=0,y(0)=2,y^'(0)=-1 is y=2e^{(2t)/3}-(7e^{(2t)/3}t)/(3)

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2.25y^{''}-3y^'+y=0,y(0)=2,y^'(0)=-1

2.25 y^{^{''}} - 3 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = - 1

y = 2 e^{\frac{2 t}{3}} - \frac{7 e ^{\frac{2 t}{3}} t}{3}

Solution steps

2.25 y^{''} (t) - 3 y^{'} (t) + y = 0

Solve linear ODE:

y = 2 e^{\frac{2 t}{3}} - \frac{7 e ^{\frac{2 t}{3}} t}{3}

y = 2 e^{\frac{2 t}{3}} - \frac{7 e ^{\frac{2 t}{3}} t}{3}

12 x^{^{''}} + 6 x^{^{'}} + 18 x = 0

b \cdot y = (a - v^{2}) y^{^{''}}

y^{^{''}} = 9 y^{^{'}} - 20 y

y^{^{''}} - 4 y^{^{''}} - 5 y = 0

0.25 x^{^{''}} + 2 x^{^{'}} + 2 x = 0

What is the solution for 2.25y^{''}-3y^'+y=0,y(0)=2,y^'(0)=-1 ?
The solution for 2.25y^{''}-3y^'+y=0,y(0)=2,y^'(0)=-1 is y=2e^{(2t)/3}-(7e^{(2t)/3}t)/(3)