Solve y'' - 2y' + 2y = 0 with y(0) = 0 and y'(0) = 2, step by step

Femtobytes

femtobytes.com/solve/ode-second-order-constant/1_-2_2_0_2

The problem

Solve this starting value problem.

y'' - 2 y' + 2 y = 0, \qquad y(0) = 0, \quad y'(0) = 2

The result

y = 2 e^{x} \sin\left( x \right)

The characteristic equation has a complex pair. Checked exactly in the original equation.

checked 7 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 2Step 1: See what the equation is asking for, and make an inspired guess

The solution. It passes through the starting point with the starting slope, shown by the short straight stroke, and from then on the equation decides everything.

See what the equation is asking for, and make an inspired guess

Why this step

The equation says: a fixed combination of $y$ , its slope $y'$ and its curvature $y''$ must cancel to zero at EVERY $x$ . The simplest way for three functions to cancel like that everywhere is for all three to be constant multiples of ONE function, so that the equation only has to balance the multiples. So we look for a function whose derivatives are multiples of itself, and the functions that do that are the exponentials $e^{rx}$ . The guess is not a trick pulled from a hat, and it costs nothing: a theorem about linear equations (Lebl, Notes on Diffy Qs, Theorem 2.1.3) says that once two independent solutions $y_1$ and $y_2$ are known, EVERY solution is $C_1 y_1 + C_2 y_2$ . So if the guess produces two, there are no others left to find.

y'' - 2 y' + 2 y = 0, \qquad y(0) = 0, \quad y'(0) = 2

y'

is the first derivative of

y

(its slope) and

y''

the second (how the slope itself changes). The two starting values will single out one solution.

The solution. It passes through the starting point with the starting slope, shown by the short straight stroke, and from then on the equation decides everything.

Turn the differential equation into an ordinary equation

Why this step

Put $y = e^{rx}$ into the equation with $r$ still unknown. Each differentiation simply brings down a factor of $r$ , so every term contains the same $e^{rx}$ , which can be taken out. What is left is an equation about the NUMBER $r$ , with no derivatives in it at all. Calculus has been traded for algebra.

y = e^{rx}, \qquad y' = r e^{rx}, \qquad y'' = r^2 e^{rx}

The derivative of

e^{rx}

r e^{rx}

(the chain rule brings down the

r

), and differentiating again brings down another.

r^{2} e^{r x} - 2 r e^{r x} + 2 e^{r x} = e^{r x} \left( r^{2} - 2 r + 2 \right)

Put these into the equation and take out the common factor

e^{rx}

. (Evaluated at several values of

r

and

x

.) checked

r^{2} - 2 r + 2 = 0

An exponential is never zero, so the product can vanish only if the bracket does. This is the characteristic equation.

The solution. It passes through the starting point with the starting slope, shown by the short straight stroke, and from then on the equation decides everything.

Second order equation with constant coefficients

See what the equation is asking for, and make an inspired guess

Turn the differential equation into an ordinary equation

Solve the characteristic equation

Turn the complex exponentials into real waves

Use the starting values to find the constants

State the solution and check it in the equation