Question

a differential equation is an equation involving an unknown function and its derivatives. consider the differential equation y(t)+y(t)=0

a. show that y = a sin t satisfies the equation for any constant a.

b. show that y = b cos t satisfies the equation for any constant b.

c. show that y = a sin t + b cos t satisfies the equation for any constants a and b.

a. find y(t) for y = a sin t. choose the correct answer below.

oa. y(t)=a cos t

ob. y(t)= - a cos t

oc. y(t)=a sin t

od. y(t)= - a sin t

Explanation:

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

Using the derivative formula $\frac{d}{dt}(\sin t)=\cos t$, we have $y^\prime(t)=\frac{d}{dt}(A\sin t)=A\cos t$.

Step2: Find second - derivative of $y = A\sin t$

Using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$, we get $y^{\prime\prime}(t)=\frac{d}{dt}(A\cos t)=-A\sin t$.

Step3: Substitute into the differential equation

Substitute $y = A\sin t$ and $y^{\prime\prime}(t)=-A\sin t$ into $y^{\prime\prime}(t)+y(t)=0$. We have $-A\sin t+A\sin t = 0$. So $y = A\sin t$ satisfies the equation.

For part b:

Step1: Find first - derivative of $y = B\cos t$

Using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$, we have $y^\prime(t)=\frac{d}{dt}(B\cos t)=-B\sin t$.

Step2: Find second - derivative of $y = B\cos t$

Using the derivative formula $\frac{d}{dt}(\sin t)=\cos t$, we get $y^{\prime\prime}(t)=\frac{d}{dt}(-B\sin t)=-B\cos t$.

Step3: Substitute into the differential equation

Substitute $y = B\cos t$ and $y^{\prime\prime}(t)=-B\cos t$ into $y^{\prime\prime}(t)+y(t)=0$. We have $-B\cos t + B\cos t=0$. So $y = B\cos t$ satisfies the equation.

For part c:

Step1: Find first - derivative of $y = A\sin t + B\cos t$

Using the sum - rule of derivatives and the basic derivative formulas, $y^\prime(t)=\frac{d}{dt}(A\sin t + B\cos t)=A\cos t - B\sin t$.

Step2: Find second - derivative of $y = A\sin t + B\cos t$

$y^{\prime\prime}(t)=\frac{d}{dt}(A\cos t - B\sin t)=-A\sin t - B\cos t$.

Step3: Substitute into the differential equation

Substitute $y = A\sin t + B\cos t$ and $y^{\prime\prime}(t)=-A\sin t - B\cos t$ into $y^{\prime\prime}(t)+y(t)=0$. We have $(-A\sin t - B\cos t)+(A\sin t + B\cos t)=0$.

For the multiple - choice part:
The second - derivative of $y = A\sin t$ is $y^{\prime\prime}(t)=-A\sin t$. So the answer is D. $y^{\prime\prime}(t)=-A\sin t$.

Answer:

a. Shown above.
b. Shown above.
c. Shown above.
Multiple - choice: D. $y^{\prime\prime}(t)=-A\sin t$