Higher-order linear equations · Definition
Superposition principle
Before you start you need to know
It turns a few known solutions into a whole family of solutions, and is why linear equations are solved by assembling pieces.
Definition
It states that if \(y_1\) and \(y_2\) are solutions of a linear homogeneous equation \(L[y]=0\), every linear combination of them with constants \(c_1,c_2\) is also a solution.
Here \(L[y]\) is the left-hand side of the equation, for example \(L[y]=y''+py'+qy\). The key is that \(L\) is linear: \(L[c_1y_1+c_2y_2]=c_1L[y_1]+c_2L[y_2]\), a property inherited from the derivative. If \(L[y_1]=L[y_2]=0\), the result is \(0\), and the combination is still a solution.
The principle fails in two situations. If the equation is nonhomogeneous, \(y=0\) is not a solution and a multiple of a solution drifts from the right-hand side: \(y=1\) solves \(y''+y=1\), but \(y=2\) gives \(2\). If it is nonlinear, like \(y'=y^2\), it does not work either. In contrast, the difference of two solutions of \(L[y]=g\) does solve the homogeneous equation.
With \(e^{2t}\) and \(e^{3t}\) in \(y''-5y'+6y=0\), every function \(c_1e^{2t}+c_2e^{3t}\) is a solution, and the two constants let us adjust \(y(0)\) and \(y'(0)\). What remains is to know when two solutions suffice to generate all of them; the Wronskian decides that.
Think about it
If \(y_1\) and \(y_2\) solve \(y''+y=0\), does \(y_1y_2\) solve that equation?
In general no: superposition covers linear combinations, not products. For \(y_1=y_2=\sin t\), \((\sin^2t)''+\sin^2t=2\cos2t+\sin^2t\neq0\).
Exercises 10
- If \(e^{t}\) and \(e^{-t}\) solve \(y''-y=0\), which of these functions also solves it?
- If \(\cos 2t\) and \(\sin 2t\) solve \(y''+4y=0\), which of these functions is a solution?
- If \(e^{2t}\) and \(e^{3t}\) solve \(y''-5y'+6y=0\), what is the value at \(t=0\) of the solution \(e^{2t}+4e^{3t}\)?
- Why is \(y=0\) always a solution of a linear homogeneous equation?
- If \(e^{t}\) and \(e^{-2t}\) solve \(y''+y'-2y=0\), which value of \(c\) makes \(ce^{t}+e^{-2t}\) equal \(3\) at \(t=0\)?
- If \(e^{-t}\) and \(e^{-2t}\) solve \(y''+3y'+2y=0\), what is \(y'(0)\) for \(y=2e^{-t}+e^{-2t}\)?
- \(y=1\) solves \(y''+y=1\). Does \(y=2\) solve that same equation?
- \(y=-1/t\) solves \(y'=y^2\), but \(y=-2/t\) does not. Why does superposition fail?
- \(y_1\) and \(y_2\) solve \(L[y]=g\) with \(g\neq0\). Which equation does \(y_1-y_2\) solve?
- If \(y_1\) solves \(L[y]=0\) and \(c\) is a constant, what is \(L[cy_1]\)?
What comes next
References
- William E. Boyce, Richard C. DiPrima, Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 12th edition, Wiley, 2021. §3.2
- Dennis G. Zill, A First Course in Differential Equations with Modeling Applications, 11th edition, Cengage, 2018. §4.1