Q30E

Question

Superposition Principle. Let $y_{1}$ be a solution to $y'' (t) + p (t) y' (t) + q (t) y (t) = g_{1} (t)$ on the interval I and let $y_{2}$ be a solution to $y'' (t) + p (t) y' (t) + q (t) y (t) = g_{2} (t)$ on the same interval. Show that for any constants $k_{1}$ and $k_{2}$ , the function $k_{1} y_{1} + k_{2} y_{2}$ is a solution on I to $y'' (t) + p (t) y' (t) + q (t) y (t) = k_{1} g_{1} (t) + k_{2} g_{2} (t)$ .

Step-by-Step Solution

Verified

Answer

$k_{1} y_{1} + k_{2} y_{2}$ is the solution to the $y'' (t) + p (t) y' (t) + q (t) y (t) = k_{1} g_{1} (t) + k_{2} g_{2} (t)$ .

1Step 1: Check whether or be the solution

If $y_{1}$ be the solution of the differential equation $y'' (t) + p (t) y' (t) + q (t) y (t) = g_{1} (t)$

Then $y_{1}^{} {'' (t) + p (t) y}_{1}^{}' (t) + q (t) y_{1} (t) = g_{1} (t)$ and if $y_{2}$ be the solution of the differential equation;

$y'' (t) + p (t) y' (t) + q (t) y (t) = g_{2} (t)$

Then $y_{2}^{} {'' (t) + p (t) y}_{2}^{}' (t) + q (t) y_{2} (t) = g_{2} (t)$

So, let's take $y = {ky}_{1} + {ky}_{2}$ then $y' = k_{1} y_{1}^{}' + k_{2} y_{2}^{}' y'' = k_{1} y_{1}^{}'' + k_{2} y_{2}^{}''$

2Step 2: Substitute the values

Substitute these equations in the differential equation:

$y'' (t) + p (t) y' (t) + q (t) y (t) = k_{1} y_{1}'' + k_{2} y_{2}'' + p (t) (k_{1} y_{1}^{}' + k_{2} y_{2}^{}') + q (t) ({ky}_{1} + {ky}_{2}) = k_{1} (y_{1}'' (t) + p (t) y_{1}' (t) + q (t) y_{1} (t)) + k_{2} (y_{2}'' (t) + p (t) y_{2}' (t) + q (t) y_{2} (t))$

Then from (1) and (2)

$y'' (t) + p (t) y' (t) + q (t) y (t) = k_{1} g_{1} (t) + k_{2} g_{2} (t)$

Therefore $k_{1} y_{1} + k_{2} y_{2}$ is the solutions to the $y'' (t) + p (t) y' (t) + q (t) y (t) = k_{1} g_{1} (t) + k_{2} g_{2} (t)$

Q29E

Q31E

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