Taylor Series Solutions to DE's

FP3 - Ex1C Q.9

y(dy/dx) = x cosh(y)

Find the series solution in ascending powers of x up to the x³ term.

I get the answer :

y = 1 + cosh(1).x²/2 + sinh(1).x³/6 + ...

but the book gives the answer:

y = 1 + cosh(1).x²/2 + ...

suggesting that the coefficient of the x³ term is 0.

Can anyone help?

Reply 1

Jonny W

8

I think the book is right.

y y' = x cosh(y)
Putting in x = 0 and y = 1 gives y' = 0.

(y')^2 + y y'' = cosh(y) + x y' sinh(y)
Putting in x = 0, y = 1 and y' = 0 gives y'' = cosh(1).

2y'y'' + y' y'' + y y''' = y' sinh(y) + y' sinh(y) + x(y'' sinh(y) + (y')^2 cosh(y))
3y'y'' + y y''' = 2y' sinh(y) + x(y'' sinh(y) + (y')^2 cosh(y))
Putting in x = 0, y = 1, y' = 0 and y'' = cosh(1) gives y''' = 0.

Reply 2

meef cheese

OP

10

Jonny W

I think the book is right.

y y' = x cosh(y)
Putting in x = 0 and y = 1 gives y' = 0.

(y')^2 + y y'' = cosh(y) + x y' sinh(y)
Putting in x = 0, y = 1 and y' = 0 gives y'' = cosh(1).

2y'y'' + y' y'' + y y''' = y' sinh(y) + y' sinh(y) + x(y'' sinh(y) + (y')^2 cosh(y))
3y'y'' + y y''' = 2y' sinh(y) + x(y'' sinh(y) + (y')^2 cosh(y))
Putting in x = 0, y = 1, y' = 0 and y'' = cosh(1) gives y''' = 0.

I got exactly the same as you, except I forgot that term. Thanks.

Taylor Series Solutions to DE's

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