Higher-order linear equations · Definition
Variation of parameters
Before you start you need to know
It solves the case that undetermined coefficients cannot reach: any continuous \(g\), in exchange for knowing two homogeneous solutions first.
Definition
A method that builds a particular solution \(y_p\) of \(y''+py'+qy=g(t)\) from two homogeneous solutions \(y_1,y_2\) and their Wronskian \(W\), by letting the constants vary.
If \(y_1,y_2\) form a fundamental set of \(y''+py'+qy=0\), the idea is to replace the constants in \(c_1y_1+c_2y_2\) by functions: \(y_p=u_1y_1+u_2y_2\). Since there are two unknown functions, one also imposes \(u_1'y_1+u_2'y_2=0\), and substituting into the equation gives \(u_1'y_1'+u_2'y_2'=g\). This is a \(2\times2\) system whose determinant is the Wronskian \(W\neq0\); by Cramer, \(u_1'=-y_2g/W\) and \(u_2'=y_1g/W\), and integrating yields the formula above.
The equation must be in standard form, with coefficient \(1\) on \(y''\): in \(2y''+2y=8t\) divide first, so \(g=4t\), not \(8t\). The constants of integration are omitted, because they would only add homogeneous solutions. With \(y''+y=\sec t\), \(y_1=\cos t\), \(y_2=\sin t\) and \(W=1\), one gets \(u_1=\ln\cos t\) and \(u_2=t\), so \(y_p=\cos t\ln\cos t+t\sin t\); undetermined coefficients would not have worked here.
The costs are that \(y_1,y_2\) are needed beforehand (for example from the characteristic equation) and the integrals can be hard; when \(g\) is a polynomial, exponential or trigonometric function, undetermined coefficients is usually faster. The general solution is always \(y=c_1y_1+c_2y_2+y_p\). The method also extends to order \(n\) with an \(n\times n\) system whose determinant is the Wronskian of \(n\) functions.
Think about it
In \(y''+y=\tan t\), why does guessing \(y_p=A\tan t\) not work?
Because \(\tan t\) is not of a type closed under differentiation: its derivatives generate \(\sec^2t\), \(\sec^2t\tan t\), etc., so there is no finite set of coefficients to adjust. Variation of parameters is used instead.
Exercises 10
- Compute \(W(y_1,y_2)\) for \(y_1=e^{t}\), \(y_2=e^{-t}\), solutions of \(y''-y=0\).
- Compute \(W(\cos2t,\sin2t)\), with \(y_1=\cos2t\), \(y_2=\sin2t\) solutions of \(y''+4y=0\).
- To apply the formula to \(2y''+2y=8t\) you must put it in the form \(y''+py'+qy=g(t)\). What is \(g(t)\)?
- When computing \(u_1\) and \(u_2\) by integration, why can the constants of integration be omitted?
- In which equation does variation of parameters work but undetermined coefficients do not?
- In \(y''+y=\tan t\) with \(y_1=\cos t\), \(y_2=\sin t\) (\(W=1\)), what is \(u_2'=\dfrac{y_1g}{W}\)?
- With \(y_1=e^{t}\), \(y_2=e^{-t}\) and \(W=-2\), find \(y_p\) for \(y''-y=e^{2t}\) by variation of parameters.
- In \(y''+y=t\) with \(y_1=\cos t\), \(y_2=\sin t\) (\(W=1\)), compute \(u_2=\int y_1g\,dt\) (no constant).
- With \(y_1=e^{t}\), \(y_2=te^{t}\) (\(W=e^{2t}\)), find \(y_p\) for \(y''-2y'+y=\dfrac{e^{t}}{t}\), \(t>0\), taking \(u_1\) and \(u_2\) with no constant.
- Find a particular solution of \(y''+y=\sec t\), \(0<t<\pi/2\), with \(y_1=\cos t\), \(y_2=\sin t\).
Related
References
- William E. Boyce, Richard C. DiPrima, Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 12th edition, Wiley, 2021. §3.6
- Dennis G. Zill, A First Course in Differential Equations with Modeling Applications, 11th edition, Cengage, 2018. §4.6