What is the solution for y^{''}=7y^'-12y,y^'(0)=1 ?

The solution for y^{''}=7y^'-12y,y^'(0)=1 is y=(e^{4t}(1-3c_{2}))/4+c_{2}e^{3t}

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y^{''}=7y^'-12y,y^'(0)=1

y^{^{''}} = 7 y^{^{'}} - 12 y, y^{^{'}} (0) = 1

y = \frac{e ^{4 t} ( 1 - 3 c _{2} )}{4} + c_{2} e^{3 t}

Solution steps

y^{''} (t) = 7 y^{'} (t) - 12 y

Solve linear ODE:

y = \frac{e ^{4 t} ( 1 - 3 c _{2} )}{4} + c_{2} e^{3 t}

y = \frac{e ^{4 t} ( 1 - 3 c _{2} )}{4} + c_{2} e^{3 t}

x^{^{''}} + 8 x^{^{'}} + 15 x = 0

y^{^{''}} + 16 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = - 2

y^{^{'''}} + 7 y^{^{''}} - 17 y^{^{'}} + 9 y = 0

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

x^{^{''}} + 3 x^{^{'}} + 2 x = 0, x^{^{'}} (0) = 2

What is the solution for y^{''}=7y^'-12y,y^'(0)=1 ?
The solution for y^{''}=7y^'-12y,y^'(0)=1 is y=(e^{4t}(1-3c_{2}))/4+c_{2}e^{3t}