Differential Equations Cauchy-Euler Equations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

To solve this we will make this variable substitution: A Cauchy-Euler equation is a specific type of D.E. that can be put in the form: This one is 2nd-order, but the solution method we will see generalizes to higher order equations as well. To solve this we will make this variable substitution: In the original equation, y is a function of t. When we substitue, we are thinking of y as a function of x, which in turn is a function of t. So it is very much like solving an integral by substitution, where you choose an intermediate variable that simplifies the integrand. It will help if we use the Leibniz notation for our derivatives. Our equation is: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

We need to find the derivatives with the chain rule, since we now have a function y(x(t)): We are using this substitution: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Next, plug in our expressions to get the equation in terms of x rather than t: Now we have an equation with constant coefficients. We know how to solve this! Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s do the homogeneous equation first. Depending on the values of α and ß we will get 2 roots, or a repeated root. For the distinct root case, call them r1 and r2 We get two solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

We get two solutions. Now we can substitute back into these solutions to get our functions of t. Next we would find a particular solution and then add to the homogeneous to get the general, as usual. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Below is an alternate method for finding the homogeneous solution. 𝑡 2 𝑦 ′′ +𝛼𝑡 𝑦 ′ +𝛽𝑦=𝑔(𝑡) 𝑡 2 𝑦 ′′ +𝛼𝑡 𝑦 ′ +𝛽𝑦=0 We will solve the homogeneous equation. It turns out that the solutions to this type of equation always work out to be power functions. We can assume a solution of the type y=tr, and solve for r. 𝑦= 𝑡 𝑟 𝑦′= 𝑟𝑡 (𝑟−1) 𝑦′′= 𝑟(𝑟−1)𝑡 (𝑟−2) 𝑡 2 𝑟(𝑟−1)𝑡 (𝑟−2) +𝛼𝑡 𝑟𝑡 (𝑟−1) +𝛽 𝑡 𝑟 =0 𝑟 2 + 𝛼−1 𝑟+𝛽=0 𝑟= −(𝛼−1)± 𝛼−1 2 −4𝛽 2 So we get 2 solutions, and the homogeneous solution is a linear combination: 𝑦 ℎ = 𝑐 1 𝑡 𝑟 1 + 𝑐 2 𝑡 𝑟 2 Next we would find a particular solution and then add to the homogeneous to get the general, as usual. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s see this method in action in a real equation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s see this method in action in a real equation. After the substitution We have: Remember, y is now a function of x here. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s see this method in action in a real equation. After the substitution We have: Remember, y is now a function of x here. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s see this method in action in a real equation. After the substitution We have: Remember, y is now a function of x here. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑡 2 𝑦 ′′ +3𝑡 𝑦 ′ +𝑦= 𝑡 −1 𝑦 ′′ +2 𝑦 ′ +𝑦= 𝑒 −𝑥 Another example: 𝑡 2 𝑦 ′′ +3𝑡 𝑦 ′ +𝑦= 𝑡 −1 After substituting x=ln(t) the equation becomes 𝑦 ′′ +2 𝑦 ′ +𝑦= 𝑒 −𝑥 Solving the homogeneous equation leads to 𝑦 𝑥 = 𝑐 1 𝑒 −𝑥 + 𝑐 2 𝑥𝑒 −𝑥 The right-hand side matches the homogeneous solution, so we get 𝑦 𝑝 𝑥 = 𝑐 2 𝑥 2 𝑒 −𝑥 We need to plug in to find the value of c3 𝑦 𝑝 ′ 𝑥 = 𝑐 2 ( 2𝑥−𝑥 2 ) 𝑒 −𝑥 𝑦 𝑝 ′′ 𝑥 = 𝑐 2 ( 2−4𝑥+𝑥 2 ) 𝑒 −𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑐 2 ( 2−4𝑥+𝑥 2 ) 𝑒 −𝑥 +2 𝑐 2 ( 2𝑥−𝑥 2 ) 𝑒 −𝑥 + 𝑐 2 𝑥 2 𝑒 −𝑥 = 𝑒 −𝑥 Another example: 𝑡 2 𝑦 ′′ +3𝑡 𝑦 ′ +𝑦= 𝑡 −1 After substituting x=ln(t) the equation becomes 𝑦 ′′ +2 𝑦 ′ +𝑦= 𝑒 −𝑥 Plugging yp and its derivatives in yields 𝑐 2 ( 2−4𝑥+𝑥 2 ) 𝑒 −𝑥 +2 𝑐 2 ( 2𝑥−𝑥 2 ) 𝑒 −𝑥 + 𝑐 2 𝑥 2 𝑒 −𝑥 = 𝑒 −𝑥 After simplifiying we have 2 𝑐 2 𝑒 −𝑥 = 𝑒 −𝑥 → 𝑐 2 = 1 2 𝑦 𝑝 𝑥 = 1 2 𝑥 2 𝑒 −𝑥 The full solution (in terms of x) is thus 𝑦 𝑥 = 𝑐 1 𝑒 −𝑥 + 𝑐 2 𝑥𝑒 −𝑥 + 1 2 𝑥 2 𝑒 −𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑦 𝑡 = 𝑐 1 𝑒 −ln⁡(𝑡) + 𝑐 2 ln⁡(𝑡)𝑒 −ln⁡(𝑡) + 1 2 ( ln 𝑡 ) 2 𝑒 −ln⁡(𝑡) Another example: 𝑡 2 𝑦 ′′ +3𝑡 𝑦 ′ +𝑦= 𝑡 −1 To get the solution in terms of y we substitute x=ln(t) again: 𝑦 𝑡 = 𝑐 1 𝑒 −ln⁡(𝑡) + 𝑐 2 ln⁡(𝑡)𝑒 −ln⁡(𝑡) + 1 2 ( ln 𝑡 ) 2 𝑒 −ln⁡(𝑡) This simplifies to 𝑦 𝑡 = 𝑐 1 𝑡 −1 + 𝑐 2 ln⁡(𝑡) 𝑡 −1 + 1 2 ( ln 𝑡 ) 2 𝑡 −1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑡 2 𝑦 ′′ −3𝑡 𝑦 ′ +4𝑦= 𝑡 2 ∙ln⁡(𝑡) 𝑦 ′′ −4 𝑦 ′ +4𝑦=𝑥 𝑒 2𝑥 Yet another example: 𝑡 2 𝑦 ′′ −3𝑡 𝑦 ′ +4𝑦= 𝑡 2 ∙ln⁡(𝑡) After substituting x=ln(t) the equation becomes 𝑦 ′′ −4 𝑦 ′ +4𝑦=𝑥 𝑒 2𝑥 Homogeneous solution is 𝑦 ℎ 𝑥 = 𝑐 1 𝑒 2𝑥 + 𝑐 2 𝑥𝑒 2𝑥 Putting this in terms of y gives 𝑦 ℎ 𝑡 = 𝑐 1 𝑡 2 + 𝑐 2 𝑡 2 ln⁡(𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑢 1 = − 𝑡 2 ln⁡(𝑡) ∙ln⁡(𝑡) 𝑡 3 𝑑𝑡=− (ln⁡(𝑡)) 3 3 Yet another example: 𝑡 2 𝑦 ′′ −3𝑡 𝑦 ′ +4𝑦= 𝑡 2 ∙ln⁡(𝑡) I’ll use variation of parameters to find the particular solution. First divide by t2 to get the equation in standard form. 𝑦 ′′ −3 𝑡 −1 𝑦 ′ +4 𝑡 −2 𝑦=ln⁡(𝑡) From the homogeneous solution we have 𝑦 1 = 𝑡 2 ; 𝑦 2 = 𝑡 2 ln 𝑡 →W 𝑦 1 , 𝑦 2 = 𝑡 3 Variation of parameters gives us a particular solution: 𝑦 𝑝 = 𝑢 1 𝑦 1 + 𝑢 2 𝑦 2 𝑢 1 = − 𝑡 2 ln⁡(𝑡) ∙ln⁡(𝑡) 𝑡 3 𝑑𝑡=− (ln⁡(𝑡)) 3 3 𝑢 2 = 𝑡 2 ∙ln⁡(𝑡) 𝑡 3 𝑑𝑡= (ln⁡(𝑡)) 2 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑦 𝑡 = 𝑐 1 𝑡 2 + 𝑐 2 𝑡 2 ln⁡(𝑡)+ 1 6 𝑡 2 (ln⁡(𝑡)) 3 Yet another example: 𝑡 2 𝑦 ′′ −3𝑡 𝑦 ′ +4𝑦= 𝑡 2 ∙ln⁡(𝑡) Solution: 𝑦 𝑡 = 𝑐 1 𝑡 2 + 𝑐 2 𝑡 2 ln⁡(𝑡)+ 1 6 𝑡 2 (ln⁡(𝑡)) 3 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB