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Section 2.3, #127: Show that the function $y=2\sin(x)+3$ satisfies the differential equation $y''+y=3$.
Solution: First compute
$$y'=2\cos(x)$$
and
$$y''=-2\sin(x).$$
Then
$$y''+y=(-2\sin(x))+(2\sin(x)+3)= 0+3=3,$$
as was to be shown.
Section 2.4, #117:
a.) Show that the derivative of an odd function is even. That is, if $f(-x)=-f(x)$, then $f'(-x)=f'(x)$.
b.) Show that the derivative of an even function is odd. That is, if $f(-x)=f(x)$, then $f'(-x)=-f'(x)$.
Solution:
a.) Use the chain rule to compute the derivative on each side of the equation $f(-x)=-f(x)$ to get
$$f'(-x) \dfrac{\mathrm{d}}{\mathrm{dx}}[-x] =-f'(x),$$
and so
$$-f'(-x)=-f'(x),$$
and so
$$f'(-x)=f'(x),$$
meaning that $f'$ is an even function, as was to be shown.
b.) Use the chain rule to compute the derivative on each side of the equation $f(-x)=f(x)$ to get
$$f'(-x) \dfrac{\mathrm{d}}{\mathrm{d}x}[-x] = f'(x),$$
so
$$-f'(-x)=f'(x),$$
or equivalently
$$f'(-x)=-f'(x),$$
showing that $f'$ is odd, as was to be shown.
