Answer to Question #131377 in Quantitative Methods for Mukesh

Question

Solve the differential equation dy/dx=(x-y)/(x+y),with y(0)=1 numerically for x=0.1 by Eular's method(five step)

Answer 1

$$Solution$$

Let's solve the differential equation by Eular's method. We had the initial value problem:

$$\frac{dy}{dx}=\frac{x-y}{x+y}\\ y(0)=1$$

Step 1

We'll start at the point $(x_0,y_0)=(0,1)$ and use step size of $h=0.01$ and proceed for 10 steps. That is, we'll approximate the solution from $t=0\ to\ t=0.1$ for our differential equation. We'll finish with a set of points that represent the solution, numerically.

We already know the first value, when $x_0=0$ which is $y_0=1$ (initial value)

We now calculate the value of the derivative at this initial point. (This tells us the direction to move.)

$\frac{dy}{dx}=f(0,1)=\frac{-1}{1}=-1$

This means the slope of the line from t=0 to t=0.1 is approximately -1

Step 2

Now, for the second step, ( Since $h=0.01$, the next point is $x+h=0+0.01=0.01$ ), we substitute what we know into Euler's Method formula, and we have:

$y(x+h) \approx y(x)+hf(x,y)\\ y_1 \approx y_0+hf(x_0,y_0)\\ y_1=y(0.01) \approx 1+0.01(-1)=0.99\\$

This means the approximate value of the solution when $x=0.01\ is\ 0.99$

We'll need the new slope at this point, so we'll know where to head next.

$\frac{dy}{dx}=f(0.01,0.99)=\frac{-0.98}{1}=-0.98$

This means the slope of the approximation line from $x=0.01\ to\ x=0.02\ is\ -0.98$. So it's a little bit steeper than the first slope we found.

Step 3

Now we are trying to find the solution value when $x=0.02$. We substitute our known values:

$y(x+h) \approx y(x)+hf(x,y)\\ y_2 \approx y_1+hf(x_1,y_1)\\ y_2=y(0.02) \approx 0.99+0.01(-0.98)=0.9802\\$

We'll need the new slope at this point, so we'll know where to head next.

$\frac{dy}{dx}=f(0.02,0.9802)=\frac{-0.9602}{1.0002}=-0.96$

This means the slope of the approximation line from $x=0.02\ to\ x=0.03\ is\ -0.96$. So it's a little bit steep than the first 2 slopes we found.

Step 4

Now we are trying to find the solution value when $x=0.03$. We substitute our known values:

$y(x+h) \approx y(x)+hf(x,y)\\ y_3 \approx y_2+hf(x_2,y_2)\\ y_3=y(0.03) \approx 0.9802+0.01(-0.96)=0.9706\\$

We'll need the new slope at this point, so we'll know where to head next.

$\frac{dy}{dx}=f(0.03,0.9706)=\frac{-0.9406}{1.0006}=-0.94$

This means the slope of the approximation line from $x=0.03\ to\ x=0.04\ is\ -0.94$. So it's a little bit steep than the first 3 slopes we found.

Step 5

We continue to calculate the subsequent steps to $x=0.1$

We present all the values up to $x=0.1$ in the following table.

(There's no final $\frac{dy}{dx}$ ​ value because we don't need it. We've found all the required $y$ values.)

We have obtained that the approximation at $x=0.1, \ y=0.909$

The differential equation $\frac{dy}{dx}=\frac{x-y}{x+y}$ at $x=0.1$ is $0.909--->Answer$