Math

Question Find the function y(x)y(x) that satisfies the first-order differential equation y=ysinxy' = y \sin x, where y=πecosxy = \pi e^{-\cos x} is a solution.

Studdy Solution

STEP 1

Assumptions
1. We are given the function y=πecosxy=\pi e^{-\cos x}.
2. We need to verify that y=ysinxy' = y \sin x.
3. The derivative of yy with respect to xx is denoted by yy'.
4. We will use the chain rule and product rule for differentiation.

STEP 2

First, we need to find the derivative of yy with respect to xx. The function yy is a product of a constant π\pi and an exponential function ecosxe^{-\cos x}. We will use the chain rule to differentiate the exponential function.
y=ddx(πecosx)y' = \frac{d}{dx} (\pi e^{-\cos x})

STEP 3

Apply the constant multiple rule to take the constant π\pi outside the derivative.
y=πddx(ecosx)y' = \pi \frac{d}{dx} (e^{-\cos x})

STEP 4

Now apply the chain rule to differentiate ecosxe^{-\cos x}. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
y=π(ecosxddx(cosx))y' = \pi \left(e^{-\cos x} \frac{d}{dx}(-\cos x)\right)

STEP 5

Differentiate cosx-\cos x with respect to xx.
ddx(cosx)=sinx\frac{d}{dx}(-\cos x) = \sin x

STEP 6

Substitute the derivative of cosx-\cos x back into the expression for yy'.
y=πecosxsinxy' = \pi e^{-\cos x} \sin x

STEP 7

Recognize that πecosx\pi e^{-\cos x} is the original function yy.
y=ysinxy' = y \sin x

STEP 8

We have shown that the derivative of yy with respect to xx is equal to ysinxy \sin x, which verifies that the given function y=πecosxy=\pi e^{-\cos x} solves the differential equation y=ysinxy' = y \sin x.
The solution has been verified.

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