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Suppose $y_1$ and $y_2$ are solutions to the differential equation: $$ y''-2y'+y=f(t)$$. Which of the following is true?

I. If $y_1(0)=y_2(0)$, then $y_1\equiv y_2$ II. If $a+b=1$, then $ay_1+by_2$ is another solution to the equation.

I know the solution to a non-homogeneous differential equation is the general solution plus one particular solution, but I'm quit confused by the relation between those particular solutions. For this problem, why is I. wrong and II. right? I can work out the general solution: $ y=C_1e^x+C_2xe^x$. Can I get $ y=C_1e^x+C_2xe^x+y_1=C_3e^x+C_4xe^x+y_2 $?

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I. is wrong because to determine both constants of a solution, $C_1, C_2$, you need two initial conditions: $y(t_0)$ and $y'(t_0)$. Then, if $y_1'(0)\neq y_2'(0)$, the second constant you determine will be different in $y_1$ and $y_2$.

II. is right. To see you must only substitute $y=ay_1 +y_2$ in the differential equation and use that the homogeneous part of both solutions will vanish when you substitute them and the linear combination of the nonhomogeneous parts will be too a particular solution because the $a+b=1$. \begin{equation} \begin{cases} y_1 = y_{h1} + y_{p1}\\ y_2 = y_{h2} + y_{p2} \end{cases} \implies y = ay_1 + by_2 = ay_{h1} + by_{h2} + ay_{p1} + by_{p2} \end{equation} Let's see if it satisfies the differential equation: \begin{equation} ay''_{h1} + by''_{h2} + ay''_{p1} + by''_{p2} - 2a'y_{h1} - 2by'_{h2} - 2ay'_{p1} - 2by'_{p2} ay_{h1} + by_{h2} + ay_{p1} + by_{p2} = (\ast) \end{equation} As $y_{h1}$ and $y_{h2}$ are solutions of the homogeneous equation, \begin{equation} \begin{cases} y_{h1}''-2y_{h1}'+y_{h1}= 0\\ y_{h2}''-2y_{h2}'+y_{h2}= 0 \end{cases} \end{equation} and \begin{equation} (\ast) = a\left(y''_{p1} - 2y'_{p1} + y_{p1}\right) + b\left(y''_{p2} - 2y'_{p2} + y_{p2}\right) = af(t) + bf(t) = (a+b)f(t) = f(t). \end{equation} In the last equality is used the hypothesis $a+b = 1$.

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