Q. 58 - Calculus

Question

Exercises $53-58$, use Euler’s method with the given $∆x$ to approximate four additional points on the graph of the solution $y\left(x\right)$. Use these points to sketch a piecewise-linear approximation of the solution.

$\frac{dy}{dx}=x^2-y,y\left(1\right)=0;∆x=0.25$.

Step-by-Step Solution

Answer

A graph for the piecewise-linear approximation of the solution for $\frac{dy}{dx}=x^2-y$ is,

Step 1 . Given information

$\frac{dy}{dx}=x^2-y,y\left(1\right)=0;∆x=0.25$.

Step 2 . The aim is to approximate the solution of the differential equation d y d x = x 2 - y that passes through the point x 0 , y 0 = 1 , 0 .

Use Euler's method to find four more points in the sequence by use of the iterative formula.

$\left(x_{k+1},y_{k+1}\right)=\left(x_k+∆x,y_k+∆y_k\right)$

Here $g\left(x,y\right)=x^2-y$ and $k=0,1,2,3$. So, first find the four additional points by using above iterative formula.

$$\begin{gathered} \left(x_0,y_0\right)=\left(1,0\right) \\ \left(x_1,y_1\right)=\left(x_0+∆x,y_0+g\left(x_0,y_0\right)∆x\right) \\ =\left(1.25,0+1\left(0.25\right)\right) \\ =\left(1.25,0.25\right) \\ \left(x_2,y_2\right)=\left(x_1+∆x,y_1+g\left(x_1,y_1\right)∆x\right) \\ =\left(1.5,0.25+1.3125\left(0.25\right)\right) \\ =\left(1.5,0.5781\right) \\ \left(x_3,y_3\right)=\left(x_2+∆x,y_2+g\left(x_2,y_2\right)∆x\right) \\ =\left(1.75,0.5781+1.6719\left(0.25\right)\right) \\ =\left(1.75,0.996\right) \\ \left(x_4,y_4\right)=\left(x_3+∆x,y_3+g\left(x_3,y_3\right)∆x\right) \\ =\left(2,0.996+2.0665\left(0.25\right)\right) \\ =\left(2,1.5126\right) \end{gathered}$$

Step 3 . Now plot the obtained five points and join the points with line segments as given below.

The approximate solution of the initial-value problem is shown by the graph by joining the line segments.