What is the solution for y^{''}+5y^'-3y=0 ?

The solution for y^{''}+5y^'-3y=0 is y=c_{1}e^{((-5+sqrt(37))t)/2}+c_{2}e^{((-5-sqrt(37))t)/2}

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y^{''}+5y^'-3y=0

y^{^{''}} + 5 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{( - 5 + 37 ) t}{2}} + c_{2} e^{\frac{( - 5 - 37 ) t}{2}}

Solution steps

y^{''} (t) + 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 5 + 37 ) t}{2}} + c_{2} e^{\frac{( - 5 - 37 ) t}{2}}

y = c_{1} e^{\frac{( - 5 + 37 ) t}{2}} + c_{2} e^{\frac{( - 5 - 37 ) t}{2}}

a^{2} \cdot \frac{d ^{2} y}{d x ^{2}} + a \cdot \frac{d y}{d x} - c \cdot y = 0

y^{^{'''}} - 11 y^{^{''}} + 24 y^{^{'}} + 36 y = 0, y (0) = 0

y^{^{''''}} + 8 y^{^{''}} - 4 y^{^{'''}} - 16 y^{^{'}} + 16 y = 0

10 y^{^{''}} + 30 y^{^{'}} + 20 y = 0

\frac{d ^{2} y}{d x ^{2}} - 14 \frac{d y}{d x} + 49 y = 0

What is the solution for y^{''}+5y^'-3y=0 ?
The solution for y^{''}+5y^'-3y=0 is y=c_{1}e^{((-5+sqrt(37))t)/2}+c_{2}e^{((-5-sqrt(37))t)/2}