MATH 308 Lecture 13

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Lecture Notes

Wronskian

$W(y_1,y_2)=y_1{y_2}^{\prime }-y_2{y_1}^{\prime }$

$W(y_1,y_2)\ne 0$ iff $y_1$ and $y_2$ form a fundamental set of solutions. Any solution is of form $y=c_1{y}_1+c_2{y}_2$

Abel's Theorem

$W(y_1,y_2)(t)=C{\mathrm{e}}^{-\int p(t)\mathrm{d}t}$

Exercise 14

$y^{″}+p(x)y^{\prime }+q(t)y=0$

$y_1$ and $y_2$ are two solutions, and they reach the maximum at the same point $t_0$. Are they a fundamental set of solutions?

Evaluate $W(y_1,y_2)(t_0)=y_1(t_0){y_2}^{\prime }(t_0)-y_2(t_0){y_1}^{\prime }(t_0)=y_1(t_0)\cdot 0-y_2(t_0)\cdot 0=0$.

Therefore, since $W(y_1,y_2)=0$, $y_1$ and $y_2$ cannot be a fundamental set of solutions.

Chapter 3.4: Reduction of Order method

Given linear, homogeneous differential equation and solution $y_1$:

Find solutions of the form $y(x)=\lambda (x)y_1(x)$

Differentiate $y$ as many times as needed:

$\begin{array}{rl}{y}^{\prime }& ={\lambda }^{\prime }{y}_1+\lambda {y_1}^{\prime }\\ y^{″}& ={\lambda }^{″}{y}_1+2{\lambda }^{\prime }{y}_1+\lambda {y_1}^{″}\\ y^{(n)}& =\dots \end{array}$

Plug the derivatives into the original differential equation. The final term should always be $0\lambda$.

Solve the reduced order differential equation for ${\lambda }^{\prime }$ and plug back into original $y$.

Exercise 15

$y^{‴}-2y^{″}-5y^{\prime }+6y=0$

We are told that $e^x$ is a solution, but we want to find more solutions of the form $y=\lambda (x){\mathrm{e}}^x$, where the ${\mathrm{e}}^x$ is the solution.

$\begin{array}{rl}{y}^{\prime }(x)& ={\lambda }^{\prime }(x){\mathrm{e}}^x+\lambda (x){\mathrm{e}}^x\\ y^{″}(x)& ={\lambda }^{″}(x){\mathrm{e}}^x+2{\lambda }^{\prime }(x){\mathrm{e}}^x+\lambda (x){\mathrm{e}}^x\\ y^{‴}(x)& ={\lambda }^{‴}(x){\mathrm{e}}^x+3{\lambda }^{″}(x){\mathrm{e}}^x,+3{\lambda }^{\prime }(x){\mathrm{e}}^x+\lambda (x){\mathrm{e}}^x\end{array}$

Therefore, our differential equation becomes

${\lambda }^{‴}(x){\mathrm{e}}^x+3{\lambda }^{″}(x){\mathrm{e}}^x+3{\lambda }^{\prime }(x){\mathrm{e}}^x+\lambda (x){\mathrm{e}}^x+2({\lambda }^{″}(x){\mathrm{e}}^x+2{\lambda }^{\prime }(x){\mathrm{e}}^x+\lambda (x){\mathrm{e}}^x)-5({\lambda }^{\prime }(x){\mathrm{e}}^x+\lambda (x){\mathrm{e}}^x)+5\lambda (x){\mathrm{e}}^x=0$

This can be reduced to

$\begin{array}{rl}{\lambda }^{‴}(x){\mathrm{e}}^x+{\lambda }^{″}{\mathrm{e}}^x-6{\lambda }^{\prime }(x){\mathrm{e}}^x+0\lambda & =0\\ {\lambda }^{‴}(x)+{\lambda }^{″}(x)-6{\lambda }^{\prime }(x)& =0\end{array}$

The $0\lambda$ is always the case. Notice that ${\lambda }^{\prime }(x)$ is a solution to a second order differential equation. Let $y(x)={\lambda }^{\prime }(x)$:

$y^{″}+y^{\prime }-6y=0$

Characteristic equation is $r^2+r-6=0$, roots are $r\in \{-3,2\}$.

$\begin{array}{rl}{\lambda }^{\prime }& =c_1{\mathrm{e}}^{-3t}+c_2{\mathrm{e}}^{2t}\\ \lambda & =-\frac{c_1}{3}{\mathrm{e}}^{-3t}+\frac{c_2}{2}{\mathrm{e}}^{2x}+C\end{array}$

Therefore, our solution is

$y=\lambda (x){\mathrm{e}}^x=(-\frac{c_1}{3}){\mathrm{e}}^{-2x}+\left(\frac{c_2}{2}\right){\mathrm{e}}^{3x}+C{\mathrm{e}}^x$

Remark

$y^{″}-4y^{\prime }+4y=0$

Characteristic equation has root $r=2$ with multiplicity 2.

$y_1={\mathrm{e}}^{2x}$

Find second solution $y_2=\lambda (x){\mathrm{e}}^{2x}$

$\begin{array}{rl}{y_2}^{\prime }& ={\lambda }^{\prime }(x){\mathrm{e}}^{2x}+2{\lambda }^{\prime }(x){\mathrm{e}}^{2x}\\ {y_2}^{″}& ={\lambda }^{″}(x){\mathrm{e}}^{2x}+4{\lambda }^{\prime }(x){\mathrm{e}}^{2x}+4\lambda (x){\mathrm{e}}^{2x}\end{array}$

$\begin{array}{rl}0& ={y_2}^{″}-4{y_2}^{\prime }+y_2\\ & =({\lambda }^{″}(x){\mathrm{e}}^{2x}+4{\lambda }^{\prime }(x){\mathrm{e}}^{2x}+4\lambda (x){\mathrm{e}}^{2x})+4({\lambda }^{\prime }(x){\mathrm{e}}^{2x}+2{\lambda }^{\prime }(x){\mathrm{e}}^{2x})+4\lambda (x){\mathrm{e}}^{2x}\end{array}$

Therefore ${\lambda }^{″}(x){\mathrm{e}}^{2x}=0$, giving $\lambda (x)=ax+b$