Question

Solve dy/dx = 2x-y with initial conditions x0 = 0, y0 = 0 by taylor’s series method. Find the approximate value of y for x = 0.2 , 0.4. Compare your results with exact values​

Answer 1

SOLUTION

TO SOLVE

$\displaystyle \sf{ \frac{dy}{dx} = 2x - y \: \: \: with \: initial \: condition \: x_0 = 0 \: and \: y_0 = 0 }$

by Taylor's series method.

Find the approximate value of y for x = 0.2 , 0.4.

Compare your results with exact values

EVALUATION

Here the given equation is

$\displaystyle \sf{ \frac{dy}{dx} = 2x - y}$

Thus we get

$\displaystyle \sf{y'= 2x - y}$

Differentiating both sides with respect to x successively we get

$\displaystyle \sf{y''= 2 - y'}$

$\displaystyle \sf{y''' = - y''}$

$\displaystyle \sf{{y}^{iv} = - y'''}$

$\sf{We \: now \: use \: \: x_0 = 0 ,y_0 = 0}$

$\displaystyle \sf{y'(0) =0 - 0 = 0}$

$\displaystyle \sf{y''(0) = 2 - y'(0) = 2 - 0 = 2}$

$\displaystyle \sf{y'''(0) = - y''(0) = - 2}$

$\displaystyle \sf{ {y}^{iv}(0) = - y'''(0) = 2}$

Putting these value in Taylor's Series we get

$\displaystyle \sf{y(x) = y(0) + xy'(0) + \frac{ {x}^{2} }{2!}y''(0) + \frac{ {x}^{3} }{3!}y'''(0) + \frac{ {x}^{4} }{4!} {y}^{iv} (0) + ...}$

$\displaystyle \sf{ \implies \: y(x) = 0 + x.0+ \frac{ {x}^{2} }{2!}.2 + \frac{ {x}^{3} }{3!}.( - 2) + \frac{ {x}^{4} }{4!}.2 + ...}$

$\displaystyle \sf{ \implies \: y(x) = {x}^{2} - \frac{ {x}^{3} }{3}+ \frac{ {x}^{4} }{12} + ...}$

$\boxed{ \: \: \displaystyle \sf{ \: y(x) = {x}^{2} - \frac{ {x}^{3} }{3}+ \frac{ {x}^{4} }{12} + ...} \: \: \: }$

Putting x = 0.2 we get

$\displaystyle \sf{ \: y(0.2) = {(0.2)}^{2} - \frac{ {(0.2)}^{3} }{3}+ \frac{ {(0.2)}^{4} }{12} + ...}$

$\displaystyle \sf{ \implies \: \: y(0.2) = 0.04 - 0.0027 + 0.0001 + ...}$

$\displaystyle \sf{ \implies \: \: y(0.2) = 0.0374}$

$\boxed{ \: \: \displaystyle \sf{ \: \: y(0.2) = 0.0374} \: }$

Putting x = 0.4 we get

$\displaystyle \sf{ \: y(0.4) = {(0.4)}^{2} - \frac{ {(0.4)}^{3} }{3}+ \frac{ {(0.4)}^{4} }{12} + ...}$

$\displaystyle \sf{ \implies \: \: y(0.4) = 0.16 - 0.0213 + 0.0021 + ...}$

$\displaystyle \sf{ \implies \: \: y(0.4) = 0.1408}$

$\boxed{ \: \: \displaystyle \sf{ \: y(0.4) = 0.1408} \: \: }$

VERIFICATION

The given differential equation is

$\displaystyle \sf{ \frac{dy}{dx} = 2x - y}$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} + y = 2x }$

$\displaystyle \sf{ Integrating \: Factor = {e}^{ \int 1.x} = {e}^{x} }$

Multiplying both sides by Integrating Factor and on integration we get

$\displaystyle \sf{ \: y. {e}^{x} = \int \: 2x. {e}^{x} dx }$

$\displaystyle \sf{ \implies \: y. {e}^{x} =2 \int \: x. {e}^{x} dx }$

$\displaystyle \sf{ \implies \: y. {e}^{x} =2 ( \: x {e}^{x} - {e}^{x} ) + c}$

$\displaystyle \sf{ \implies \: y =2 ( \: x - 1 ) + c. {e}^{ - x}}$

Where C is integration constant

Now y = 0 when x = 0 gives

$\displaystyle \sf{ \implies \: 0 =2 ( \: 0 - 1 ) + c. {e}^{ - 0}}$

$\displaystyle \sf{ \implies \:c = 2}$

So the solution is

$\boxed{ \: \: \displaystyle \sf{ y (x)=2 ( \: x - 1 ) + 2. {e}^{ - x}} \: \: }$

Putting x = 0.2 we get

$\displaystyle \sf{ y(0.2) =2 ( \: 0.2 - 1 ) + 2. {e}^{ - 0.2}}$

$\displaystyle \sf{ \implies y(0.2) = - 1.6 + 2 \times 0.8187}$

$\displaystyle \sf{ \implies y(0.2) =0.0374}$

$\boxed{ \: \: \displaystyle \sf{ \: \: y(0.2) = 0.0374} \: }$

Putting x = 0.4 we get

$\displaystyle \sf{ y (0.4)=2 ( 0.4 - 1 ) + 2. {e}^{ - 0.4}}$

$\displaystyle \sf{ \implies y (0.4)= - 1.2 + 1.3408}$

$\displaystyle \sf{ \implies \: y(0.4) = 0.1408} \: \:$

$\boxed{ \: \: \displaystyle \sf{ \: y(0.4) = 0.1408} \: \: }$

Hence verified

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