Question

Use taylors series method to find approximate Value of y at x =0.2,given that dy/dx =2y+3ex,y(0)=0 comapre the result by solving the differential equation

Answer 1

We are given; $\displaystyle y\prime=2y+3e^x,\ y(0)=0,\ h=0.1$ (since step length is not given, I assume a step length of $0.1$), $\displaystyle x_0=0,\text{and } y_0=0$.

Using Taylor's series method, we are to find $y(0.2).$

Now, differentiating the given DE successively yields;

$\displaystyle y\prime=2y+3e^x\\ y\prime\prime=2y\prime+3e^x\\ y\prime\prime\prime=2y\prime\prime+3e^x\\$

Now, substituting yields;

$\displaystyle y_0\prime=2y_0+3e^{x_0}=3\\ y_0\prime\prime=2y_0\prime+3e^{x_0}=9\\ y_0\prime\prime\prime=2y_0\prime\prime+3e^{x_0}=21\\$

Putting these values into the Taylor's series yields;

$\displaystyle y(0.1)=y_1=y_0+hy_0\prime+\frac{h^2}{2!}y_0\prime\prime+\frac{h^3}{3!}y_0\prime\prime\prime+\cdots$

$\displaystyle =0+0.1(3)+\frac{0.1^2}{2}\times9+\frac{0.1^3}{6}\times21+\cdots=0.3485$

Next, $x_1=0.1$, and $y_1=0.3485$. Thus substituting yields;

$\displaystyle y_1\prime=2y_1+3e^{x_1}=4.012512754\\ y_1\prime\prime=2y_1\prime+3e^{x_1}=11.34053826\\ y_1\prime\prime\prime=2y_1\prime\prime+3e^{x_1}=25.99658928\\$

Putting these values into the Taylor's series yields;

$\displaystyle y(0.2)=y_2=y_1+hy_1\prime+\frac{h^2}{2!}y_1\prime\prime+\frac{h^3}{3!}y_1\prime\prime\prime+\cdots$

$\displaystyle =0.3485+0.1(4.012512754)+\frac{0.1^2}{2}\times11.34053826$

$\displaystyle +\frac{0.1^3}{6}\times25.99658928+\cdots=0.8107867316$

Thus, the approximate value of $y(0.2)$ using Taylor's series is;

$y(0.2)=0.8107867316$

Solving directly yields;

$\displaystyle y\prime=2y+3e^x\Rightarrow y\prime-2y=3e^x$

Integrating Factor is;

$\displaystyle e^{\int(-2)\ dx}=e^{-2x}$

$\displaystyle$$\displaystyle (y\prime-2y)e^{-2x}=3e^x\times e^{-2x}\\ \Rightarrow \frac{d}{dx}(ye^{-2x})=3e^{-x}\\ \Rightarrow ye^{-2x}=\int3e^{-x}\ dx=3\int e^{-x}\ dx=-3e^{-x}+c$, where c is an arbitrary constant

$\displaystyle \Rightarrow y=ce^{2x}-3e^{x}$

Since $\displaystyle y(0)=0, \text{ then }c=3\Rightarrow y=3e^{2x}-3e^{x}\\$

Now,

$\displaystyle y(0.2)=3e^{2\times0.2}-3e^{0.2}=3e^{0.4}-3e^{0.2}=0.8112658184$