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Find 2 values of m so that the function y=tm is a solution to the 2nd order linear ODE (assume t>0):

ty′′+2y′=0

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Final answer:

To find the values of m so that the function y=tm is a solution to the given second-order linear ODE, we use the characteristic equation. The two values of m are -1 + i√3 and -1 - i√3.

Step-by-step explanation:

In order for the function y=tm to be a solution to the given second-order linear ordinary differential equation (ODE), we need to substitute this function into the equation and solve for the variable m.

Substituting y=tm into the equation ty′′+2y′=0, we get t(m''t+m')+2mt'=0. Simplifying this equation, we have m'' + 2m' + 2m = 0. This is a linear homogeneous ODE.

To find the values of m, we can use the characteristic equation. Assuming m=e^rt, where r is a constant, we substitute m into the homogeneous ODE and solve for r. The characteristic equation is r^2 + 2r + 2 = 0.

Using the quadratic formula, the roots of the characteristic equation are r = -1 ± i√3. Since m=e^rt, we can write the general solution as m = c1 * e^((-1+i√3)t) + c2 * e^((-1-i√3)t), where c1 and c2 are constants.

So, the two values of m are -1 + i√3 and -1 - i√3.

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