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Chris L T521 · #8 May 27th, 2008, 01:59 AM

Technique #3: Variation of Parameters:

We have dealt with non-homogeneous equations where $f(x)$ had the form of $x^n \text{,} \ x^ne^{\alpha x} \text{,} \ x^ne^{\alpha x}\sin(\beta x) \text{,} \ etc$ . However, what if $f(x)$ was $\sec x$ ? $\sin^{-1}x$ ? $\ln x$ ? We would now need a new technique to conquer these non-homogeneous equations. That technique is known as variation of parameters.

If we have a differential equation in the form $\frac{d^2y}{dx^2}+P(x)\frac{dy}{dx}+Q(x)y=f(x)$ , we want $y_p=u_1(x)y_1+u_2(x)y_2$ . But how do we find $u_1(x)$ and $u_2(x)$ ?? We will substitute $y_p$ into the differential equation. This will give us the following:

$u_1^{//} y_1 + u_1^/ y_1^/ + u_1^/ y_1^/ + u_1 y_1^{//} + u_2^{//} y_2 + u_2^/ y_2^/$ $+ u_2^/ y_2^/ + u_2 y_2^{//} + P(x)\left[ {u_1^/ y_1 + u_1 y_1^/ + u_2^/ y_2 + u_2 y_2^/ } \right]$ $+ Q(x)\left[ {u_1 y_1 + u_2 y_2 } \right]$
$= u_1 \left[ {y_1^{//} + P(x)y_1^/ + Q(x)y_1 } \right] + u_2 \left[ {y_2^{//} + P(x)y_2^/ + Q(x)y_2 } \right]$ $+ u_1^{//} y_1 + u_1^/ y_1^/ + u_2^{//} y_2 + u_2^/ y_2^/ + P(x)\left[ {u_1^/ y_1 + + u_2^/ y_2 } \right]$ $+ u_1^/ y_1^/ + u_2^/y_2^/$
$= \frac{d}{{dx}}\left[ {u_1^/ y_1 + u_2^/ y_2 } \right] + P(x)\left[ {u_1^/ y_1 + + u_2^/ y_2 } \right] + u_1^/ y_1^/ + u_2^/ y_2^/$
$=f(x)$

Since we seek $u_1$ and $u_2$ , we need two equations. To get these equations, we need to impose 2 conditions onto $u_1$ and $u_2$ The first condition is that $L\left[y_p\right]=f(x)$ . The second one we can apply is not given, thus, we need to come up with one.

Recall that $y_p^/ = u_1^/ y_1 + u_1 y_1^/ + u_2^/ y_2 + u_2 y_2^/$ . Rearranging, we get $y_p^/ = (u_1 y_1^/ + u_2 y_2^/ ) + (u_1^/ y_1 + u_2^/ y_2 )$ .

To avoid the appearance of the second derivatives $u_1^{//}$ and $u_2^{//}$ , we impose the second condition that the second sum must equal zero:

$u_1^/ y_1 + u_2^/ y_2 =0$

With these assumptions, the DE becomes $u_1^/ y_1^/ + u_2^/ y_2^/ = f(x)$ .

Now, we have a system of equations:

$\left\{\begin{array}{l}u_1^/ y_1 + u_2^/ y_2 =0 \\u_1^/ y_1^/ + u_2^/ y_2^/ = f(x)\\\end{array}\right.$

To solve this, we will use Cramer's Rule:

If Ax = b is a system of n linear equations in n unknowns such that $\det (A) \ne 0$ , then the system has a unique solution. This solution is
$x_1 = \frac{{\det (A_1 )}}{{\det (A)}}$ ,..., $x_2 = \frac{{\det (A_2 )}}{{\det (A)}}$ ,..., $x_n= \frac{{\det (A_n )}}{{\det (A)}}$

where $A_j$ is the matrix obtained by replacing the terms in the $j^{th}$ column of A by the entries in matrix b.

To apply Cramer's Rule, we write the system in matrix form:

$\left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {u_1^/ } \\ {u_2^/ } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 \\ {f(x)} \\ \end{array} } \right]$

Thus, according to Cramer's rule,

$u_1^/ = \frac{{\det \left[ {\begin{array}{*{20}c} 0 & {y_2 } \\ {f(x)} & {y_2^/ } \\ \end{array} } \right]}}{{\det \left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]}}$ and $u_2^/ = \frac{{\det \left[ {\begin{array}{*{20}c} {y_1 } & 0 \\ {y_1^/ } & {f(x)} \\ \end{array} } \right]}}{{\det \left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]}}$

We represent the determinant matrices as $W_1$ , $W_2$ , and $W$ . Particularly,

$W_1^{} = \det \left[ {\begin{array}{*{20}c} 0 & {y_2 } \\ {f(x)} & {y_2^/ } \\ \end{array} } \right]$ , $W_2^{} = \det \left[ {\begin{array}{*{20}c} {y_1 } & 0 \\ {y_1^/ } & {f(x)} \\ \end{array} } \right]$ , and $W = \det \left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]$

We recongnize $W$ as the Wronskian of $y_1$ and $y_2$ . Due to the linear independence of $y_1$ and $y_2$ , $W(y_1,y_2)\neq 0$ .

We can now find $u_1$ and $u_2$ .

Example 19:

Solve $y''+y=\tan x$ .

Solve the homogeneous equation:

$y''+y=0 \implies r^2+1=0 \implies r=\pm i$

Thus, $y_c=c_1\cos(x)+c_2\sin(x)$ .

Now that we have $y_1=\cos(x)$ and $y_2=\sin(x)$ we can use variation of parameters.

$W_1=\det \left[ {\begin{array}{*{20}c} 0 & {\sin x } \\ {\tan x} & {\cos x} \\ \end{array} } \right]=-\frac{\sin^2(x)}{cos(x)}=\frac{\cos^2(x)-1}{cos(x)}=\cos(x)-\sec(x)$

$W_2=\det \left[ {\begin{array}{*{20}c} {cos(x) } & 0 \\ {-sin(x) } & {\tan(x)} \\ \end{array} } \right]=\cos(x)\tan(x)=\sin(x)$

$W = \det \left[ {\begin{array}{*{20}c} {\cos(x)} & {\sin(x)} \\ {-\sin(x)} & {\cos(x)} \\ \end{array} } \right]=\cos^2(x)-\left(-\sin^2(x)\right)=cos^2(x)+sin^2(x)=1$

Therefore $u_1^/=\frac{W_1}{W}=\cos(x)-\sec(x)$ and $u_2^/=\frac{W_2}{W}=\sin(x)$ .

Now find $u_1$ and $u_2$ .

$u_1=\int \left(\cos(x)-\sec(x)\right)\,dx=\sin(x)-\ln\left|\sec(x)+\tan(x)\right|$

$u_2=\int \sin(x)\,dx=-\cos(x)$

Therefore, our particular solution will be:

$\color{red}\boxed{y=\left[\sin(x)-\ln\left|\sec(x)+\tan(x)\right|\right]\cos(x)-\frac{1}{2}\sin(2x)}$

My next couple posts will be on applications of non-homogeneous differential equations (in particular, the spring mass systems)...

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