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### Variation of Parameters

#### Discussion

The method of undetermined coefficients is the most efficient way to solve an inhomogeneous equation if we can guess the proper form for the particular solution. Unfortunately we can't always guess the proper form. If we can find the general solution to the homogeneous equation, the method of variation of parameters will always work. Because the method is complicated, we will only cover the second order case. Consider the following example $$ y''+y=\tan(x) $$ We first find two linearly independent solutions to the homogeneous equation $y''+y=0$. In this case two linearly independent solutions are $\cos(x)$ and $\sin(x)$. We now guess that the solution to the inhomogeneous equation will take the form $$ y(x)=u(x)\cos(x)+v(x)\sin(x) $$ Differentiating this yields $$ y'(x)=-u(x)\sin(x)+v(x)\cos(x)+u'(x)\cos(x)+v'(x)\sin(x) $$ We should now differentiate this again to compute $y''$ but this is getting ugly. $y''$ will have 6 terms. At this point I observe I have 2 degrees of freedom in my guess for the solution (i.e. 2 unknown functions) and I only need to satisfy one equation. With 2 unknown functions I should be able to satisfy 2 equations and I will now tack on the additional condition that besides solving the original problem, $y''+y=\tan(x)$, I will choose $u$ and $v$ so that $$ u'(x)\cos(x)+v'(x)\sin(x)=0. $$ Using this I can then get $$ y'(x)=-u(x)\sin(x)+v(x)\cos(x) $$ and I then differentiate again to get $$ y''(x)=-u(x)\cos(x)-v(x)\sin(x)-u'(x)\sin(x)+v'(x)\cos(x) $$ So plugging into $y''+y=\tan(x)$ we get $$ y''+y= \eqalign {&-u(x)\cos(x)-v(x)\sin(x)-u'(x)\sin(x)+v'(x)\cos(x) \cr &\qquad\qquad+u(x)\cos(x) + v(x)\sin(x)\cr} \buildrel{\text{set}}\over{=} \tan(x) $$ From which we get the equation $$ -u'(x)\sin(x)+v'(x)\cos(x)=\tan(x) $$ We now have two equations for $u'(x)$ and $v'(x)$. $$ \begin{align} u'(x)\cos(x)+v'(x)\sin(x)&=0 \\ -u'(x)\sin(x)+v'(x)\cos(x)&=\tan(x) \end{align} \tag 1 $$ Multiplying the first equation by $\sin(x)$ and the second equation by $\cos(x)$ and adding we obtain $$ \begin{align} v'(x)(\sin^2(x)+\cos^2(x))&=\sin(x) \\ v'(x)&=\sin(x) \\ v(x)&=-\cos(x)+C_1 \end{align} $$ Multiplying the first equation in (1) by $\cos(x)$ and the second equation in (1) by $-\sin(x)$ and adding we obtain $$ \begin{align} u'(x)(\cos^2(x)+\sin^2(x))&=-\sin^2(x)/\cos(x) \\ u'(x)&=-\sin^2(x)/\cos(x) \\ u(x)&=\sin(x)-\log\left(\frac{\sin(x)+1}{\cos(x)}\right)+C_2 \end{align} $$ Finally we get $$ \begin{align} y(x)&=\sin(x)\cos(x)-\log\left(\frac{\sin(x)+1}{\cos(x)}\right)\cos(x) +C_2\cos(x)-\cos(x)\sin(x)+C_1\sin(x) \\ &=-\log\left(\frac{\sin(x)+1}{\cos(x)}\right)\cos(x)+C_2\cos(x)+C_1\sin(x) \end{align} $$ In this one case, I*do not*advise you to learn the procedure for solving inhomogeneous equations using variation of parameters. The manipulations given above are complicated and most students have lots of trouble with them. Using them, I can derive the following formula and I advise you to just copy down the formula on your crib sheet and use it whenever necessary.

This looks at first glance to have three arbitrary constants but in fact it only has two. Any given solution can be obtained with every choice of $x_0$ by making the right choice of $C_1$ and $C_2$. The advantage of this form is that while we often won't be able to evaluate the integrals, we will always be able to evaluate any definite integral from $x_0$ to $x_0$. So if we have an initial value problem with the values given at $x_0$, we will be able to work out the values of $C_1$ and $C_2$ even if we can't compute the integrals directly.

#### Paradigm

Solve $x^2y''+xy'-4y=x$ given that two linearly independent solutions of the corresponding homogeneous equation are $y_1(x)=x^2$ and $y_2(x)=x^{-2}$.
*Step 0:* The theorem above only applies when the coefficient of
$y''$ is 1, so we need to divide through the whole equation by $x^2$ to
get
$$
y''+(1/x)y'-(4/x^2)y=1/x
$$
*Step 1:* Find two linearly independent solutions of the
corresponding homogeneous problem.

Here we are given that two solutions are $y_1(x)=x^2$ and $y_2(x)=x^{-2}$.

*Step 2:* Evaluate the solution of the inhomogeneous problem using
the formula.
$$
W(y_1,y_2)(s)=s^2\times-2s^{-3}-2s\times s^{-2}=-4s^{-1}
$$
We don't need to worry about the lower limit in this problem and can just
treat the integrals as indefinite integrals in $x$.
$$ \begin{align}
&\int^x\frac{s^{-2}s^{-1}}{-4s^{-1}}\,ds=-\frac14\int^x s^{-2}\,ds=
\frac{x^{-1}}4
\\
&\int^x\frac{s^2s^{-1}}{-4s^{-1}}\,ds=-\frac14\int^x
s^2\,ds=-\frac{x^3}{12}
\end{align} $$
So we get
$$ \begin{align}
y(x)&=-x^2\times\frac{x^{-1}}4+x^{-2}\times-\frac{x^3}{12}+C_1x^2+C_2x^{-2}
\\
&=-\frac{x}{4}-\frac{x}{12}+C_1x^2+C_2x^{-2} \\
&=-\frac{x}{3}+C_1x^2+C_2x^{-2}
\end{align} $$

EXAMPLE 1: Solve the initial value problem $$y''+6y'+5y=\sqrt{x},\qquad y(1)=0,\quad y'(1)=0$$ FIRST: Find the general solution

Step 1: The homogeneous equation is $y''+6y'+5y=0$

SubStep 1: $(D^2+6D+5)y=(D+5)(D+1)y=0$

SubStep 2: roots are $-5$ and $-1$

SubStep 3: The general solution is $C_1e^{-5x}+C_2e^{-x}$ and $e^{-5x}$ and $e^{-x}$ are two linearly independent solutions of the homogeneous equation.

Step 2: $$ W(e^{-5x},e^{-x})=e^{-5x}\times-e^{-x}-(-5e^{-5x}\times e^{-x})=4e^{-6x} $$ We choose $x_0=1$ since we are given initial values at $1$. Then $$ \begin{align} &\int_1^x\frac{e^{-s}\sqrt{s}}{4e^{-6s}}\,ds=\int_1^x\frac{e^{5s}\sqrt s}{4} \,ds \\ &\int_1^x\frac{e^{-5s}\sqrt{s}}{4e^{-6s}}\,ds = \int_1^x\frac{e^{s}\sqrt{s}}{4}\,ds \end{align} $$ I can evaluate neither of these integrals. So the best I can do for the general solution to the inhomogeneous problem is $$ y(x)=-e^{-5x}\int_1^x\frac{e^{5s}\sqrt s}{4}\,ds +e^{-x}\int_1^x\frac{e^{s}\sqrt{s}}{4}\,ds +C_1e^{-5x}+C_2e^{-x} $$

SECOND: Plug in the initial values and solve for the constants. $$ \begin{align} y(1)&=-e^{-5}\times0+e^{-1}\times0+C_1e^{-5}+C_2e^{-1} \buildrel{\text{set}}\over{=} 0 \\ y'(x)&=5e^{-5x}\int_1^x\frac{e^{5s}\sqrt s}{4}\,ds -e^{-5x}\frac{e^{5x}\sqrt{x}}{4} \\ &\qquad\qquad-e^{-x}\int_1^x\frac{e^{s}\sqrt{s}}{4}\,ds +e^{-x}\frac{e^x\sqrt{x}}{4} \\ &\qquad\qquad-5C_1e^{-5x}-C_2e^{-x} \\ y'(1)&=5e^{-5}\times0-\frac 14 -e^{-1}\times0+\frac 14 -5C_1e^{-5}-C_2e^{-1} \buildrel{\text{set}}\over{=} 0 \end{align} $$ Here we have used the fact that the derivative of the integral is the original function. So the initial values yield the two equations $$ \begin{align} C_1e^{-5}+C_2e^{-1}&=0 \\ -5C_1e^{-5}-C_2e^{-1}&=0 \end{align} $$ These have the solution $C_1=0$ and $C_2=0$ so our solution to the initial value problem is $$y(x)=-e^{-5x}\int_1^x\frac{e^{5s}\sqrt s}{4}\,ds +e^{-x}\int_1^x\frac{e^{s}\sqrt{s}}{4}\,ds $$

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©2010, 2014 Andrew G. Bennett