Method of variation of parameters

The method of variation of parameters finds a particular solution of a nonhomogeneous linear ODE for any forcing function — including ones undetermined coefficients can’t handle (like $\tan x$, $\sec x$, $\ln x$). The trade-off is more involved integrals.

For a second-order linear ODE in standard form $y^{\prime \prime }+a(x)y^{\prime }+b(x)y=f(x)$ with two known linearly independent homogeneous solutions $y_1(x),y_2(x)$:

$y_p(x)=v_1(x)y_1(x)+v_2(x)y_2(x)$

where $v_1,v_2$ are functions (not constants — that’s the “variation”) satisfying:

$\{\begin{array}{l}{v}_1^{\prime }{y}_1+v_2^{\prime }{y}_2=0\\ v_1^{\prime }{y}_1^{\prime }+v_2^{\prime }{y}_2^{\prime }=f(x)\end{array}$

Solve this $2\times 2$ system for $v_1^{\prime },v_2^{\prime }$ via Cramer’s rule:

$v_1^{\prime }(x)=\frac{-f(x)y_2(x)}{W[y_1,y_2](x)},v_2^{\prime }(x)=\frac{f(x)y_1(x)}{W[y_1,y_2](x)}$

Then integrate to get $v_1,v_2$.

The trick

The ansatz $y_p=v_1{y}_1+v_2{y}_2$ has two unknown functions. A second-order ODE is one equation. With two unknowns and one equation, we have freedom to impose an extra constraint. The standard choice is

$v_1^{\prime }{y}_1+v_2^{\prime }{y}_2=0$

This makes $y_p^{\prime }=v_1{y}_1^{\prime }+v_2{y}_2^{\prime }$ — clean, no second-derivative-of-$v_i$ terms. Then $y_p^{\prime \prime }=v_1^{\prime }{y}_1^{\prime }+v_1{y}_1^{\prime \prime }+v_2^{\prime }{y}_2^{\prime }+v_2{y}_2^{\prime \prime }$, and substituting into the ODE (using that $y_1,y_2$ satisfy the homogeneous version) collapses to:

$v_1^{\prime }{y}_1^{\prime }+v_2^{\prime }{y}_2^{\prime }=f(x)$

So we get the system above.

Why $W\ne 0$ matters

The denominator in the Cramer’s-rule formula is the Wronskian $W[y_1,y_2]$. For linearly independent $y_1,y_2$ on the interval, $W\ne 0$ — so $v_1^{\prime },v_2^{\prime }$ are well-defined.

If $y_1,y_2$ were linearly dependent, the system has no unique solution and the method fails. This is one reason linear independence matters.

Worked example 1

$x{y}^{\prime \prime }-(x+2)y^{\prime }+2y=x^3$ on $(0,\infty )$. Standard form: $y^{\prime \prime }-\frac{x+2}{x}{y}^{\prime }+\frac{2}{x}y=x^2$.

Two known linearly independent homogeneous solutions: $y_1(x)=e^x$, $y_2(x)=x^2+2x+2$. (The PDF provides these.)

Wronskian:

$W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2=e^x(2x+2)-e^x(x^2+2x+2)=e^x(-x^2)$

So $W=-x^2{e}^x$.

Cramer’s:

$v_1^{\prime }=\frac{-f{y}_2}{W}=\frac{-x^2(x^2+2x+2)}{-x^2{e}^x}=(x^2+2x+2)e^{-x}$

$v_2^{\prime }=\frac{f{y}_1}{W}=\frac{x^2\cdot e^x}{-x^2{e}^x}=-1$

Integrate:

$v_1=\int (x^2+2x+2)e^{-x}dx$

Integration by parts gives $v_1=-(x^2+4x+6)e^{-x}$ (omitting constant).

$v_2=\int -1dx=-x$

Particular solution:

$y_p=v_1{y}_1+v_2{y}_2=-(x^2+4x+6)e^{-x}\cdot e^x+(-x)(x^2+2x+2)$

$=-(x^2+4x+6)-x(x^2+2x+2)=-x^2-4x-6-x^3-2x^2-2x$

$=-x^3-3x^2-6x-6$

General solution:

$y(x)=c_1{e}^x+c_2(x^2+2x+2)-x^3-3x^2-6x-6$

Worked example 2: split forcing

$y^{\prime \prime }+y=\tan x+3x-1$ on $(-\pi /2,\pi /2)$.

Homogeneous: $y_1=\cos x$, $y_2=\sin x$, $W=1$.

Use Superposition principle for $y_p$: solve each piece separately.

For $\tan x$: variation of parameters since undetermined coefficients can’t handle $\tan x$.

$v_1^{\prime }=-\sin x\tan x=-{\sin }^{2}x/\cos x,v_2^{\prime }=\cos x\tan x=\sin x$

Integrate: $v_1=\sin x-\ln |\sec x+\tan x|$ (after some work), $v_2=-\cos x$.

$y_{p,1}=v_1\cos x+v_2\sin x$

Substituting $v_1=\sin x-\ln |\sec x+\tan x|$ and $v_2=-\cos x$:

$y_{p,1}=\sin x\cos x-\cos x\ln |\sec x+\tan x|-\sin x\cos x=-\cos x\ln |\sec x+\tan x|$

The two $\sin x\cos x$ terms cancel exactly. (More generally: any constants of integration in $v_1,v_2$ contribute multiples of $y_1,y_2$, which are homogeneous solutions and can be absorbed into $c_1{y}_1+c_2{y}_2$ in the general solution.)

For $3x-1$: undetermined coefficients gives $y_{p,2}=3x-1$ (after computation).

Total: $y_p=-\cos x\ln |\sec x+\tan x|+3x-1$.

When it’s the right method

Choose variation of parameters when:

• The forcing function isn’t one of the “nice” types ($\tan x$, $\sec x$, $1/(1+x^2)$, etc.).
• The homogeneous solutions $y_1,y_2$ are non-elementary (e.g., Bessel functions) — variation of parameters doesn’t care, as long as you have them.
• You need a particular solution and undetermined coefficients doesn’t apply.

For ODEs with sinusoidal/exponential/polynomial forcing, Method of undetermined coefficients is faster — guess and solve for constants instead of computing integrals. But variation of parameters always works (assuming you can do the integrals).

Higher orders

For an $n$-th order linear ODE, the method generalizes: $y_p=v_1{y}_1+\cdots +v_n{y}_n$ with $n-1$ constraint equations $v_1^{\prime }{y}_1^{(k)}+\cdots =0$ for $k=0,\dots ,n-2$, plus one final equation that yields the right-hand side $f$. Cramer’s rule applies similarly using the higher-order Wronskian.

In practice, second-order is the common case in engineering courses.