Question

a differential equation is an equation involving an unknown function and its derivatives. consider the differential equation y(t)+y(t)=0. a. find y(t) for y = a sin t. choose the correct answer below. a. y(t)=-a cos t b. y(t)=-a sin t c. y(t)=a cos t d. y(t)=a sin t b. show that y = a sin t satisfies the equation for any constant a. c. show that y = b cos t satisfies the equation for any constant b. d. show that y = a sin t + b cos t satisfies the equation for any constants a and b.

Explanation:

Step1: Find the first - derivative of $y = A\sin t$

Using the derivative formula $\frac{d}{dt}(\sin t)=\cos t$, we have $y'=\frac{d}{dt}(A\sin t)=A\cos t$. Then find the second - derivative. Using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$, we get $y''=\frac{d}{dt}(A\cos t)=-A\sin t$.

Step2: Check if $y = A\sin t$ satisfies $y''(t)+y(t) = 0$

Substitute $y = A\sin t$ and $y''=-A\sin t$ into the equation $y''(t)+y(t)=0$. We have $-A\sin t+A\sin t = 0$, so $y = A\sin t$ satisfies the equation.

Step3: Check if $y = B\cos t$ satisfies $y''(t)+y(t) = 0$

First, find the first - derivative of $y = B\cos t$. Using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$, we have $y'=-B\sin t$. Then find the second - derivative. Using the derivative formula $\frac{d}{dt}(\sin t)=\cos t$, we get $y''=-B\cos t$. Substitute $y = B\cos t$ and $y''=-B\cos t$ into the equation $y''(t)+y(t)=0$. We have $-B\cos t + B\cos t=0$, so $y = B\cos t$ satisfies the equation.

Step4: Check if $y=A\sin t + B\cos t$ satisfies $y''(t)+y(t) = 0$

Find the first - derivative: $y'=A\cos t - B\sin t$. Then find the second - derivative: $y''=-A\sin t - B\cos t$. Substitute $y = A\sin t + B\cos t$ and $y''=-A\sin t - B\cos t$ into the equation $y''(t)+y(t)=0$. We get $(-A\sin t - B\cos t)+(A\sin t + B\cos t)=0$.

Answer:

a. $y''(t)=-A\sin t$ (Answer is A)
b. Since $y = B\cos t$, $y'=-B\sin t$ and $y''=-B\cos t$, and $y'' + y=-B\cos t + B\cos t = 0$, so $y = B\cos t$ satisfies the equation.
c. Since $y=A\sin t + B\cos t$, $y'=A\cos t - B\sin t$, $y''=-A\sin t - B\cos t$, and $y'' + y=(-A\sin t - B\cos t)+(A\sin t + B\cos t)=0$, so $y = A\sin t + B\cos t$ satisfies the equation.