Questions: Find the general solution to the equation (y^(4)-12 y^prime prime prime+36 y^prime prime=0). NOTE: Use (c1, c2, c3), and (c4) as arbitrary constants. [y=]

Solution Steps

To solve the differential equation \( y^{(4)} - 12 y^{\prime \prime \prime} + 36 y^{\prime \prime} = 0 \), we can use the characteristic equation method. The characteristic equation for this differential equation is \( r^4 - 12r^3 + 36r^2 = 0 \). We solve this polynomial equation to find the roots, which will help us construct the general solution.

1. Write the characteristic equation corresponding to the differential equation.
2. Solve the characteristic equation to find the roots.
3. Use the roots to write the general solution of the differential equation.

The characteristic equation derived from the differential equation \( y^{(4)} - 12 y^{\prime \prime \prime} + 36 y^{\prime \prime} = 0 \) is given by: \[ r^4 - 12r^3 + 36r^2 = 0 \]

By factoring the characteristic equation, we find the roots: \[ r^2(r^2 - 12r + 36) = 0 \] This simplifies to: \[ r^2(r - 6)^2 = 0 \] Thus, the roots are: \[ r = 0, \quad r = 6 \quad (\text{with multiplicity } 2) \]

The general solution of the differential equation can be constructed from the roots. Since we have a root \( r = 0 \) with multiplicity 1 and \( r = 6 \) with multiplicity 2, the general solution is: \[ y(t) = c_1 + c_2 t + c_3 e^{6t} + c_4 t e^{6t} \] where \( c_1, c_2, c_3, \) and \( c_4 \) are arbitrary constants.

Final Answer