Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Taylor Series method (first order differential equation)

Solution:
Given `y'=-2x-y, y(0)=-1, h=0.1, y(0.5)=?`

Here, `x_0=0,y_0=-1,h=0.1,x_n=0.5`

Differentiating successively, we get

Derivative steps

`d/(dx)(-2x-y)`

`=-d/(dx)(2x)-d/(dx)(y)`

`=-2-y'`

Now, `d^2/(dx^2)(-2x-y)=d/(dx)(-2-y')`

`=-d/(dx)(2)-d/(dx)(y')`

`=-0-y''`

`=0-y''`

`=-y''`

Now, `d^3/(dx^3)(-2x-y)=d/(dx)(-y'')`

`=-y'''`

`y'=-2x-y`

`y''=-2-y'`

`y'''=-y''`

`y^(iv)=-y'''`

Now substituting, we get
`y_0'=-2x_0-y_0=1`

`y_0''=-2-y_0'=-3`

`y_0'''=-y_0''=3`

`y_0^(iv)=-y_0'''=-3`

Putting these values in Taylor Series, we have
`y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0^(iv) + ...`

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for `n=0,x_0=0,y_0=-1`

`=-1+0.1*(1)+(0.1)^2/(2)*(-3)+(0.1)^3/(6)*(3)+(0.1)^4/(24)*(-3)+...`

`=-1+0.1-0.015+0+0+...`

`=-0.9145`

`x_1=x_0+h=0+0.1=0.1`

Now substituting, we get
`y_1'=-2x_1-y_1=0.7145`

`y_1''=-2-y_1'=-2.7145`

`y_1'''=-y_1''=2.7145`

`y_1^(iv)=-y_1'''=-2.7145`

Putting these values in Taylor Series, we have
`y_2 = y_1 + hy_1' + h^2/(2!) y_1'' + h^3/(3!) y_1''' + h^4/(4!) y_1^(iv) + ...`

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for `n=1,x_1=0.1,y_1=-0.9145`

`=-0.9145+0.1*(0.7145)+(0.1)^2/(2)*(-2.7145)+(0.1)^3/(6)*(2.7145)+(0.1)^4/(24)*(-2.7145)+...`

`=-0.9145+0.0715-0.0136+0+0+...`

`=-0.8562`

`x_2=x_1+h=0.1+0.1=0.2`

Now substituting, we get
`y_2'=-2x_2-y_2=0.4562`

`y_2''=-2-y_2'=-2.4562`

`y_2'''=-y_2''=2.4562`

`y_2^(iv)=-y_2'''=-2.4562`

Putting these values in Taylor Series, we have
`y_3 = y_2 + hy_2' + h^2/(2!) y_2'' + h^3/(3!) y_2''' + h^4/(4!) y_2^(iv) + ...`

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for `n=2,x_2=0.2,y_2=-0.8562`

`=-0.8562+0.1*(0.4562)+(0.1)^2/(2)*(-2.4562)+(0.1)^3/(6)*(2.4562)+(0.1)^4/(24)*(-2.4562)+...`

`=-0.8562+0.0456-0.0123+0+0+...`

`=-0.8225`

`x_3=x_2+h=0.2+0.1=0.3`

Now substituting, we get
`y_3'=-2x_3-y_3=0.2225`

`y_3''=-2-y_3'=-2.2225`

`y_3'''=-y_3''=2.2225`

`y_3^(iv)=-y_3'''=-2.2225`

Putting these values in Taylor Series, we have
`y_4 = y_3 + hy_3' + h^2/(2!) y_3'' + h^3/(3!) y_3''' + h^4/(4!) y_3^(iv) + ...`

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for `n=3,x_3=0.3,y_3=-0.8225`

`=-0.8225+0.1*(0.2225)+(0.1)^2/(2)*(-2.2225)+(0.1)^3/(6)*(2.2225)+(0.1)^4/(24)*(-2.2225)+...`

`=-0.8225+0.0222-0.0111+0+0+...`

`=-0.811`

`x_4=x_3+h=0.3+0.1=0.4`

Now substituting, we get
`y_4'=-2x_4-y_4=0.011`

`y_4''=-2-y_4'=-2.011`

`y_4'''=-y_4''=2.011`

`y_4^(iv)=-y_4'''=-2.011`

Putting these values in Taylor Series, we have
`y_5 = y_4 + hy_4' + h^2/(2!) y_4'' + h^3/(3!) y_4''' + h^4/(4!) y_4^(iv) + ...`

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for `n=4,x_4=0.4,y_4=-0.811`

`=-0.811+0.1*(0.011)+(0.1)^2/(2)*(-2.011)+(0.1)^3/(6)*(2.011)+(0.1)^4/(24)*(-2.011)+...`

`=-0.811+0.0011-0.0101+0+0+...`

`=-0.8196`

`x_5=x_4+h=0.4+0.1=0.5`

`:.y(0.5)=-0.8196`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n^(iv)`	`x_(n+1)`	`y_(n+1)`
0	0	-1	1	-3	3	-3	0.1	-0.9145
1	0.1	-0.9145	0.7145	-2.7145	2.7145	-2.7145	0.2	-0.8562
2	0.2	-0.8562	0.4562	-2.4562	2.4562	-2.4562	0.3	-0.8225
3	0.3	-0.8225	0.2225	-2.2225	2.2225	-2.2225	0.4	-0.811
4	0.4	-0.811	0.011	-2.011	2.011	-2.011	0.5	-0.8196

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