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Variation of parameters is a method for finding particular solutions to nonhomogeneous linear differential equations. It is especially useful when undetermined coefficients does not apply, such as equations with lnx\ln x, tanx\tan x, or variable coefficients. This cheat sheet helps students organize the setup, formulas, and solution steps for second-order and higher-order ODEs.

It is designed as a quick reference for homework, exams, and review.

Key Facts

  • For a second-order equation y+p(x)y+q(x)y=g(x)y'' + p(x)y' + q(x)y = g(x), first solve the homogeneous equation y+p(x)y+q(x)y=0y'' + p(x)y' + q(x)y = 0.
  • If y1y_1 and y2y_2 are linearly independent homogeneous solutions, then yh=C1y1+C2y2y_h = C_1y_1 + C_2y_2.
  • The Wronskian for two solutions is W(y1,y2)=y1y2y1y2W(y_1,y_2) = y_1y_2' - y_1'y_2, and variation of parameters requires W(y1,y2)0W(y_1,y_2) \neq 0 on the interval.
  • A particular solution has the form yp=u1(x)y1(x)+u2(x)y2(x)y_p = u_1(x)y_1(x) + u_2(x)y_2(x), where u1u_1 and u2u_2 are functions to be found.
  • For y+p(x)y+q(x)y=g(x)y'' + p(x)y' + q(x)y = g(x), the derivative formulas are u1=y2gWu_1' = -\frac{y_2g}{W} and u2=y1gWu_2' = \frac{y_1g}{W}.
  • After finding u1u_1' and u2u_2', integrate to get u1=y2gWdxu_1 = \int -\frac{y_2g}{W}\,dx and u2=y1gWdxu_2 = \int \frac{y_1g}{W}\,dx.
  • The general solution of a nonhomogeneous linear ODE is y=yh+ypy = y_h + y_p.
  • If the equation is not in standard form, divide by the leading coefficient so that the coefficient of yy'' is 11 before applying the formulas.

Vocabulary

Nonhomogeneous ODE
A differential equation with a nonzero forcing term, such as y+p(x)y+q(x)y=g(x)y'' + p(x)y' + q(x)y = g(x) where g(x)0g(x) \neq 0.
Homogeneous equation
The related equation formed by setting the forcing term equal to zero, such as y+p(x)y+q(x)y=0y'' + p(x)y' + q(x)y = 0.
Fundamental solutions
A set of linearly independent solutions to the homogeneous equation that can be combined to form yhy_h.
Wronskian
A determinant used to test linear independence, given by W(y1,y2)=y1y2y1y2W(y_1,y_2) = y_1y_2' - y_1'y_2 for two functions.
Particular solution
One solution ypy_p that satisfies the full nonhomogeneous differential equation.
General solution
The complete solution y=yh+ypy = y_h + y_p that combines the homogeneous solution with a particular solution.

Common Mistakes to Avoid

  • Forgetting to put the ODE in standard form is wrong because the formulas u1=y2gWu_1' = -\frac{y_2g}{W} and u2=y1gWu_2' = \frac{y_1g}{W} assume the coefficient of yy'' is 11.
  • Using dependent homogeneous solutions is wrong because W(y1,y2)=0W(y_1,y_2) = 0 makes the variation of parameters formulas invalid.
  • Dropping the negative sign in u1=y2gWu_1' = -\frac{y_2g}{W} is wrong because it changes the particular solution and usually fails when substituted back into the ODE.
  • Adding arbitrary constants to u1u_1 and u2u_2 is unnecessary because those constants only reproduce terms already included in yh=C1y1+C2y2y_h = C_1y_1 + C_2y_2.
  • Stopping after finding u1u_1' and u2u_2' is wrong because the functions u1u_1 and u2u_2 must be obtained by integration before building ypy_p.

Practice Questions

  1. 1 Use variation of parameters to solve y+y=secxy'' + y = \sec x on an interval where cosx0\cos x \neq 0.
  2. 2 For y3y+2y=e3xy'' - 3y' + 2y = e^{3x}, use y1=exy_1 = e^x and y2=e2xy_2 = e^{2x} to compute W(y1,y2)W(y_1,y_2), then set up u1u_1' and u2u_2'.
  3. 3 Find a particular solution for y+4y=tan(2x)y'' + 4y = \tan(2x) using y1=cos(2x)y_1 = \cos(2x) and y2=sin(2x)y_2 = \sin(2x).
  4. 4 Explain why variation of parameters can work for forcing terms where undetermined coefficients does not apply.