Math  /  Calculus

Question3 It is given that, at any point on the graph of y=f(x),dy dx=exsinxyy=\mathrm{f}(x), \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{e}^{x} \sin x-y. (i) Show that d3y dx3=d2y dx22y\frac{\mathrm{d}^{3} y}{\mathrm{~d} x^{3}}=\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-2 y.
The graph of y=f(x)y=\mathrm{f}(x) passes through the origin OO. (ii) Find the Maclaurin series for yy, up to and including the term in x4x^{4}. [4] (iii) Hence, find the Maclaurin series for e2xsin2x\mathrm{e}^{2 x} \sin 2 x, up to and including the term in x2x^{2}. [2]

Studdy Solution

STEP 1

What is this asking? We're given a *differential equation* that tells us how the *slope* of a function changes, and we need to find a *series representation* of the function near x=0x = 0, and then use that to find a series for another function. Watch out! Don't forget the *chain rule* when taking derivatives, especially with the exponential and trigonometric functions!
Also, keep track of your negative signs when simplifying.

STEP 2

1. Find the third derivative.
2. Find the Maclaurin series for yy.
3. Find the Maclaurin series for e2xsin2x\mathrm{e}^{2x} \sin 2x.

STEP 3

We are given dydx=exsinxy\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x \sin x - y.
Let's **differentiate** both sides with respect to xx to find the *second derivative*.
Remember the *product rule* for exsinx\mathrm{e}^x \sin x!

STEP 4

d2ydx2=exsinx+excosxdydx\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \mathrm{e}^x \sin x + \mathrm{e}^x \cos x - \frac{\mathrm{d}y}{\mathrm{d}x}

STEP 5

Now, let's **differentiate** again to find the *third derivative*.

STEP 6

d3ydx3=exsinx+excosx+excosxexsinxd2ydx2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \mathrm{e}^x \sin x + \mathrm{e}^x \cos x + \mathrm{e}^x \cos x - \mathrm{e}^x \sin x - \frac{\mathrm{d}^2y}{\mathrm{d}x^2}

STEP 7

Notice that exsinxexsinx\mathrm{e}^x \sin x - \mathrm{e}^x \sin x add to zero.
We can **simplify** this to:

STEP 8

d3ydx3=2excosxd2ydx2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 2\mathrm{e}^x \cos x - \frac{\mathrm{d}^2y}{\mathrm{d}x^2}

STEP 9

We know dydx=exsinxy\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x \sin x - y.
Substituting this into the second derivative expression gives us:

STEP 10

d2ydx2=exsinx+excosx(exsinxy)=excosx+y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \mathrm{e}^x \sin x + \mathrm{e}^x \cos x - (\mathrm{e}^x \sin x - y) = \mathrm{e}^x \cos x + y

STEP 11

Substituting this into the third derivative expression gives us:

STEP 12

d3ydx3=2excosx(excosx+y)=excosxy\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 2\mathrm{e}^x \cos x - (\mathrm{e}^x \cos x + y) = \mathrm{e}^x \cos x - y

STEP 13

Oops, that's not quite what we wanted!
Let's go back to the given differential equation and notice that exsinx=dydx+y\mathrm{e}^x \sin x = \frac{\mathrm{d}y}{\mathrm{d}x} + y.
We can substitute this into our *second derivative* expression:

STEP 14

d2ydx2=(dydx+y)+excosxdydx=y+excosx\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = (\frac{\mathrm{d}y}{\mathrm{d}x} + y) + \mathrm{e}^x \cos x - \frac{\mathrm{d}y}{\mathrm{d}x} = y + \mathrm{e}^x \cos x So excosx=d2ydx2y\mathrm{e}^x \cos x = \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - y.

STEP 15

Now, substitute this into our *third derivative* expression:

STEP 16

d3ydx3=2(d2ydx2y)d2ydx2=d2ydx22y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 2(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - y) - \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2y Yay!

STEP 17

Since the graph passes through the origin, we know y(0)=0y(0) = 0.
Using the given differential equation, we can find dydxx=0=e0sin00=0\frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x=0} = \mathrm{e}^0 \sin 0 - 0 = 0.

STEP 18

Using the expression for the second derivative, d2ydx2=y+excosx\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = y + \mathrm{e}^x \cos x, we find d2ydx2x=0=0+1=1\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\Big|_{x=0} = 0 + 1 = 1.

STEP 19

Using the expression for the third derivative, d3ydx3=d2ydx22y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2y, we find d3ydx3x=0=10=1\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\Big|_{x=0} = 1 - 0 = 1.

STEP 20

Differentiating the third derivative gives us d4ydx4=d3ydx32dydx\frac{\mathrm{d}^4y}{\mathrm{d}x^4} = \frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2\frac{\mathrm{d}y}{\mathrm{d}x}, so d4ydx4x=0=10=1\frac{\mathrm{d}^4y}{\mathrm{d}x^4}\Big|_{x=0} = 1 - 0 = 1.

STEP 21

The Maclaurin series is given by y=y(0)+xdydxx=0+x22!d2ydx2x=0+x33!d3ydx3x=0+x44!d4ydx4x=0+y = y(0) + x\frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x=0} + \frac{x^2}{2!}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\Big|_{x=0} + \frac{x^3}{3!}\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\Big|_{x=0} + \frac{x^4}{4!}\frac{\mathrm{d}^4y}{\mathrm{d}x^4}\Big|_{x=0} + \dots

STEP 22

Substituting our values, we get y=0+0+x22(1)+x36(1)+x424(1)+=x22+x36+x424+y = 0 + 0 + \frac{x^2}{2}(1) + \frac{x^3}{6}(1) + \frac{x^4}{24}(1) + \dots = \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \dots

STEP 23

We know dydx=exsinxy\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x \sin x - y.
If we substitute 2x2x for xx, we get dyd(2x)2=e2xsin2xy(2x)\frac{\mathrm{d}y}{\mathrm{d}(2x)} \cdot 2 = \mathrm{e}^{2x} \sin 2x - y(2x), or e2xsin2x=2dydxx=2x+y(2x)\mathrm{e}^{2x} \sin 2x = 2\frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x=2x} + y(2x).

STEP 24

Substituting the Maclaurin series for yy, we get e2xsin2x=2(x+4x26+)+(4x22+8x36+)=2x+4x23+2x2+=2x+10x23+\mathrm{e}^{2x} \sin 2x = 2(x + \frac{4x^2}{6} + \dots) + (\frac{4x^2}{2} + \frac{8x^3}{6} + \dots) = 2x + \frac{4x^2}{3} + 2x^2 + \dots = 2x + \frac{10x^2}{3} + \dots

STEP 25

The Maclaurin series for yy is x22+x36+x424+\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \dots.
The Maclaurin series for e2xsin2x\mathrm{e}^{2x} \sin 2x is 2x+10x23+2x + \frac{10x^2}{3} + \dots

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