Use Eulers method to approximate y1 for the solution yt to t

Use Euler\'s method to approximate y(.1) for the solution y(t) to the 2nd order IVP y\'\'-2y\'-y=0; y(0)=1, y\'(0)=2

Solution

The given Differential Equation is

y\'\' - 2 y\' - y = 0

Integrating both sides, we get

y\' - 2y - y dx + C₁ = 0

Ley y = e^x

So, y dx = e^x dx = e^x

^So, y\' - 2e^x - e^x dx + C₁ = 0

Implies that y\' - 2e^x - e^x + C₁ = 0

Implies that y\' - 3e^x  + C₁ = 0 If x = 0 y\' = 2

So 2 - 3 e⁰ + C₁ = 0 So C₁ = 1

Integrating

y - 3e^x  + C₁ x + C₂  = 0

So the equation can be written as

y(x) = 3 e^x - C₁ x - C₂

If x = 0, y = 1 So, 1 = 3e⁰ - C₁ 0 - C_{2 Implies that} C₂ = 3-1 = 2

So the equation is

y(x) = 3 e^x - x - 2

So, the approximate value of y(0.1) = 3 e^0.1 - 0.1 - 2 =

( e^x = 1 +x + x²/2! + x³/3! + .......... = 1 + 0.1 + (0.1)²2! + (0.1)/3! +.........

1 + 0.1 + 0.005 + 0.00016 +....... = 1.10516)

So, y(0.1) = 3*1.10516 -0.1 -2 = 3.31548 - 0.1 -2 = 1.21548