In this blog, I will be teaching you how you can solve a differential equation with the help of Taylors series. 

Before we could dive into solving differential equations, lets rewid some points about Taylor series.

Taylors Series Formula

\(f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\)

* The Taylor series provides an approximation of the function near the point a, and the accuracy of the approximation depends on how many terms are used.

Lets Learn How to Solve a Differential Equation using Taylors Series 

How to Solve a Differential Equation using Taylors Series 

Q. Approximate the solution of y' = x - y² , ( y = 1 when x = 0 ) . Find y when x = 0.1 using first four terms of Taylor series. 

Solution:

To approximate the solution of the differential equation \( y' = x - y^2 \), with \( y = 1 \) when \( x = 0 \), using the first four terms of the Taylor series, we can follow these steps:

* Find the first three derivatives of the function \( y(x) \). since up to four terms of Taylor series we need three derivatives.

 * Evaluate these derivatives at the point \( x = 0 \) and \( y = 1 \).

 * Substitute these values into the Taylor series formula.

 * Use the Taylor series to approximate \( y(0.1) \).


The given differential equation is \( y' = x - y^2 \).

The Taylor series expansion for \( y(x) \) around \( x = 0 \) is given by:

\(y(x) = y(0) + y'(0) x + \frac{y''(0)}{2!} x^2 + \frac{y'''(0)}{3!} x^3 + \ldots\)


First, let's find the first three derivatives of \( y(x) \) with respect to \( x \):

\(y' = x- y^2\)

\( y'' = 1 - 2y. y'\)

\( y''' = -2\{{y'}^2 + y. y''\}\)


Now, evaluate these derivatives at \( x = 0 \) and \( y = 1 \):


\(y'(0) = 0 - y(0)^2 = 0 - 1^2 = -1  \)

\(y''(0) = 1 - 2.y(0) . y'(0) = 1 - 2.1.(-1) = 1+2 = 3 \)

\(y'''(0) = -2\{y'(0)^2 +y(0). y''(0)\} = -2 \{-1^2 + 1.3\} = -8\)


Substituting these values into the Taylor series formula, we get:


\(y(x) = y(a) + y'(a)(x-a) + \frac{y''(a)}{2!}(x-a)^2 + \frac{y'''(a)}{3!}(x-a)^3 + \ldots\)

\(= y(0) + y'(0)(x-0) + \frac{y''(0)}{2!}(x-0)^2 + \frac{y'''(0)}{3!}(x-0)^3 + \ldots\)

\(= 1 + (-1).x + \frac{3}{2!}x^2 + \frac{-8}{3!}x^3 \)

\(= 1 + -x + \frac{3}{2}x^2 + \frac{-4}{3}x^3 \)

Which is the required equation.


Lets find the value of y(0.1),

Now, to approximate \( y(0.1) \), substitute \( x = 0.1 \) into the Taylor series:

\(y(0.1) = 1 + -0.1 + \frac{3}{2}0.1^2 + \frac{-4}{3}0.1^3  = 0.9136\)


Therefore, the approximate value of \( y(0.1) \) using the first four terms of the Taylor series is \( 0.9136 \).