Second-order ODE initial value problem

[Kang00]

Hello,

I have problems with a differential equation of second order:

y''-2y'+5y=0

with IV:

y(0)=2 and y'(0)=-4

I tried to slove it with aid of Laplace transform, due to the fact that both nessesarv IV are given.
The Laplace transform turned into a difficult convolution:

 $ y(s) = L^(-1) \{ \frac{2s-8}{(s-1)^2 + 4} \} $

This Problem is part of an old Exam and it should be able to slove it in a short time without any electronic calculator.

My Question is: Is there an easier way to solve an equation like that?

 

[Ishuda]

Hello,

I have problems with a differential equation of second order:

y''-2y'+5y=0

with IV:

y(0)=2 and y'(0)=-4

I tried to slove it with aid of Laplace transform, due to the fact that both nessesarv IV are given.
The Laplace transform turned into a difficult convolution:

 $ y(s) = L^(-1) \{ \frac{2s-8}{(s-1)^2 + 4} \} $

This Problem is part of an old Exam and it should be able to slove it in a short time without any electronic calculator.

My Question is: Is there an easier way to solve an equation like that?

Since the coefficients for the function are constant you can use the characteristic method. That is,
y = e^{r x}
y' = r y
y'' = r² y
and the equation becomes
(r² - 2 r + 5) y = 0
or, since y is never zero,
r² - 2 r + 5 = 0
This gives an r of r₁ and r₂ and the general solution becomes
y = A e^{r₁ x} + B e^{r₂ x}
with IV:

y(0)=2 and y'(0)=-4

 

[HallsofIvy]

Personally, I feel that "Laplace transform" is NEVER a good way to solve an ordinary linear differential equation! Instead, solve the "characteristic equation". In a "differential equations" course, that typically is taught well before "Laplace transform". Did you learn it in a physics or engineering class?

For the linear differential equation, with constant coefficients, $\displaystyle a_n y^{(n)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$ has "characteristic equation" $\displaystyle a_n t^n+ \cdot\cdot\cdot+ a_1t+ a_0= 0$.
(If you "try" $\displaystyle y= e^{tx}$ as a solution, it reduces to $\displaystyle (a_n t^n+ \cdot\cdot\cdot+ a_1t+ a_0)e^{tx}= 0a$. Since $\displaystyle e^{tx}$ is never 0, $\displaystyle a_n t^n+ \cdot\cdot\cdot+ a_1t+ a_0$ must be 0. If solutions to that characteristic equation are $\displaystyle s_1$, $\displaystyle s_2$, etc., then the solutions to the differential equation are of the form $\displaystyle Ae^{s_1x}$, $\displaystyle Be^{s_2x}$, etc. with "A", "B", etc.)

Here, your equation is $\displaystyle y''- 2y'+ 4y$ so the characteristic equation is $\displaystyle s^2- 2s+ 5= s^2- 2s+ 4+ 1= (s- 2)^2+ 1= 0$ which has roots s= 2+ i and s= 2- i. Solutions to the differential equation are $\displaystyle Ae^{(2+i)x}$ and $\displaystyle Be^{(2- i)x}$.

Since all numbers in the differential equation are real, we would prefer a solution written in terms of real numbers only. So write $\displaystyle e^{(2+i)x}= e^{2x}e^{ix}$. But now we can use the identity, $\displaystyle e^{ix}= cos(x)+ i sin(x)$. That is, $\displaystyle e^2e^{ix}= e^2(cos(x)+ i sin(x))$ so that $\displaystyle Ae^{(2+i)x)}+ Be^{(2- ix)}= Ae^{2x}(cos(x)+ i sin(x))+ Be^{2x}(cos(x)- i sin(x))= e^{2x}((A+ B) cos(x)+ i(A- B)sin(x))= e^{2x}(Ccos(x)+ Dsin(x))$ where C= A+ B, D= Ai- Bi.

Given $\displaystyle y(x)= e^{2x}(C cos(x)+ D sin(x)$, we also have $\displaystyle y'= (e^{2x})'(C cos(x)+ D sin(x))'+ (e^{2x})(C cos(x)+ D sin(x))'$$\displaystyle = 2e^{2x}(C cos(x)+ D sin(2x))+ e^{2x}(-C sin(x)+ D cos(x))$.

$\displaystyle e^0= 1$, $\displaystyle cos(0)= 1$, and $\displaystyle sin(0)= 0$ so
$\displaystyle y(0)= C= 2$ and $\displaystyle y'(0)= D= -4$.

That is, $\displaystyle y(x)= e^{2x}(2 cos(x)- 4 sin(x))$.