3.4 Repeated Roots

Methods of reduction of order:

• Consider the homogeneous variable-coefficient ODE

  • $y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$
  • supposed we already know that one solution is $y_1(t).$ then guess a second solution of the form $y_2(t)=v(t)y_1(t)$ for some function v(t) to be determined

    • Derivation of $v(t):$

      • assumption $y_1(t)$ is a sol
      • goal: find $v(t)$ $s\cdot t{y}_2(t)=v(t)y_1(t)$ is also a sol
      • $y_2=v{y}_1$

$$\begin{array}{r}\\ y_2=v{y}_1\\ y_2^{\prime }=v^{\prime }{y}_1+v{y}_1^{\prime }\\ y_2^{\prime \prime }=(v^{\prime \prime }{y}_1+v_1^{\prime }{y}_1^{\prime })+(v^{\prime }{y}_1^{\prime }+v{y}_1^{\prime \prime })⟹v^{\prime \prime }{y}_1+2v^{\prime }{y}_1^{\prime }+v{y}^{\prime \prime }\\ \text{ pluging y”, y’, and y into the orignal function }\\ v^{\prime \prime }(y_1)+v^{\prime }(2y_1^{\prime }+p(t)y_1)+v(y_1^{\prime \prime }+p(t)y_1^{\prime }+q(t)y_1)=0\\ \text{ since y1 is a solution we can just cancel out the y1 solution}\\ v^{\prime \prime }(y_1)+v^{\prime }(2y_1^{\prime }+p(t)y_1)=0\\ w(t)=v_1(t)\\ w^{\prime }(t)=v^{\prime \prime }(t)\\ w^{\prime }{y}_1+w(2y_1^{\prime }+p(t)y_1)=0\end{array}$$

Euler equations:

• it takes the form of

  • $a{t}^2{y}^{\prime \prime }+bt{y}^{\prime }+cy=0$ with constants a,b,c with a=!0

• Solving

  • to solve these you change the variable to x=lnt

Examples

• find the second solution

  • $t^2{y}^{\prime \prime }-4t{y}^{\prime }+$