Math / Calculus

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Problem 1. (1 point) Find $y$ as a function of $t$ if $\begin{array}{ll} y(0)=5, \quad y^{\prime}(0)=8 \\ y(t)= & 6 y^{\prime \prime}+33 y=0 \\ & \end{array}$
Note: This particular weBWorK problem can't handle complex numbers, so write your answer in terms of sines and cosines, rather than using e to a complex power.

Studdy Solution

STEP 1

What is this asking? We need to find a formula for $y(t)$ that satisfies the given differential equation $6y'' + 33y = 0$ and the initial conditions $y(0) = 5$ and $y'(0) = 8$ . Watch out! Don't forget to apply both initial conditions to find the specific solution.
Also, remember we're looking for a solution in terms of sines and cosines.

STEP 2

1. Solve the characteristic equation.
2. Write the general solution.
3. Apply the initial conditions.

STEP 3

Let's solve the characteristic equation!
We replace $y''$ with $r^2$ and $y$ with $1$ in the differential equation $6y'' + 33y = 0$ to get the characteristic equation $6r^2 + 33 = 0$ .

STEP 4

Now, we solve for $r$ .
We have $6r^2 = -33$ , so $r^2 = -\frac{33}{6} = -\frac{11}{2}$ .
Taking the square root of both sides gives us $r = \pm \sqrt{-\frac{11}{2}} = \pm i\sqrt{\frac{11}{2}}$ .
Since we have a pair of imaginary roots, our solution will involve sines and cosines!

STEP 5

With imaginary roots $\pm i\sqrt{\frac{11}{2}}$ , the general solution has the form $y(t) = c_1 \cos\left(\sqrt{\frac{11}{2}}t\right) + c_2 \sin\left(\sqrt{\frac{11}{2}}t\right)$ , where $c_1$ and $c_2$ are constants we need to find.

STEP 6

First, let's use $y(0) = 5$ .
Plugging in $t = 0$ into our general solution, we get $y(0) = c_1 \cos(0) + c_2 \sin(0) = c_1 \cdot 1 + c_2 \cdot 0 = c_1$ .
Since $y(0) = 5$ , we have $c_1 = \mathbf{5}$ !

STEP 7

Next, we use $y'(0) = 8$ .
Let's find the first derivative of our general solution: $y'(t) = -\sqrt{\frac{11}{2}}c_1 \sin\left(\sqrt{\frac{11}{2}}t\right) + \sqrt{\frac{11}{2}}c_2 \cos\left(\sqrt{\frac{11}{2}}t\right)$ .

STEP 8

Now, plug in $t = 0$ to get $y'(0) = -\sqrt{\frac{11}{2}}c_1 \sin(0) + \sqrt{\frac{11}{2}}c_2 \cos(0) = \sqrt{\frac{11}{2}}c_2$ .
Since $y'(0) = 8$ , we have $8 = \sqrt{\frac{11}{2}}c_2$ , so $c_2 = \mathbf{\frac{8}{\sqrt{\frac{11}{2}}}} = 8\sqrt{\frac{2}{11}}$ .

STEP 9

Substitute the values of $c_1$ and $c_2$ back into the general solution to get our particular solution: $y(t) = 5\cos\left(\sqrt{\frac{11}{2}}t\right) + 8\sqrt{\frac{2}{11}}\sin\left(\sqrt{\frac{11}{2}}t\right)$ .

STEP 10

The solution to the differential equation is $y(t) = 5\cos\left(\sqrt{\frac{11}{2}}t\right) + 8\sqrt{\frac{2}{11}}\sin\left(\sqrt{\frac{11}{2}}t\right)$ .

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Studdy Solution

STEP 1

STEP 2

1. Solve the characteristic equation.
2. Write the general solution.
3. Apply the initial conditions.

STEP 3

Let's solve the characteristic equation!
We replace $y''$ with $r^2$ and $y$ with $1$ in the differential equation $6y'' + 33y = 0$ to get the characteristic equation $6r^2 + 33 = 0$ .

STEP 4

Now, we solve for $r$ .
We have $6r^2 = -33$ , so $r^2 = -\frac{33}{6} = -\frac{11}{2}$ .
Taking the square root of both sides gives us $r = \pm \sqrt{-\frac{11}{2}} = \pm i\sqrt{\frac{11}{2}}$ .
Since we have a pair of imaginary roots, our solution will involve sines and cosines!

STEP 5

With imaginary roots $\pm i\sqrt{\frac{11}{2}}$ , the general solution has the form $y(t) = c_1 \cos\left(\sqrt{\frac{11}{2}}t\right) + c_2 \sin\left(\sqrt{\frac{11}{2}}t\right)$ , where $c_1$ and $c_2$ are constants we need to find.

STEP 6

First, let's use $y(0) = 5$ .
Plugging in $t = 0$ into our general solution, we get $y(0) = c_1 \cos(0) + c_2 \sin(0) = c_1 \cdot 1 + c_2 \cdot 0 = c_1$ .
Since $y(0) = 5$ , we have $c_1 = \mathbf{5}$ !

STEP 7

Next, we use $y'(0) = 8$ .
Let's find the first derivative of our general solution: $y'(t) = -\sqrt{\frac{11}{2}}c_1 \sin\left(\sqrt{\frac{11}{2}}t\right) + \sqrt{\frac{11}{2}}c_2 \cos\left(\sqrt{\frac{11}{2}}t\right)$ .

STEP 8

Now, plug in $t = 0$ to get $y'(0) = -\sqrt{\frac{11}{2}}c_1 \sin(0) + \sqrt{\frac{11}{2}}c_2 \cos(0) = \sqrt{\frac{11}{2}}c_2$ .
Since $y'(0) = 8$ , we have $8 = \sqrt{\frac{11}{2}}c_2$ , so $c_2 = \mathbf{\frac{8}{\sqrt{\frac{11}{2}}}} = 8\sqrt{\frac{2}{11}}$ .

STEP 9

Substitute the values of $c_1$ and $c_2$ back into the general solution to get our particular solution: $y(t) = 5\cos\left(\sqrt{\frac{11}{2}}t\right) + 8\sqrt{\frac{2}{11}}\sin\left(\sqrt{\frac{11}{2}}t\right)$ .

STEP 10

The solution to the differential equation is $y(t) = 5\cos\left(\sqrt{\frac{11}{2}}t\right) + 8\sqrt{\frac{2}{11}}\sin\left(\sqrt{\frac{11}{2}}t\right)$ .

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Studdy Solution

STEP 1

STEP 2

1. Solve the characteristic equation.2. Write the general solution.3. Apply the initial conditions.

STEP 3

Let's **solve the characteristic equation**!We replace y′′y''y′′ with r2r^2r2 and yyy with 111 in the differential equation 6y′′+33y=06y'' + 33y = 06y′′+33y=0 to get the characteristic equation 6r2+33=06r^2 + 33 = 06r2+33=0.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

STEP 10

The solution to the differential equation is y(t)=5cos⁡(112t)+8211sin⁡(112t)y(t) = 5\cos\left(\sqrt{\frac{11}{2}}t\right) + 8\sqrt{\frac{2}{11}}\sin\left(\sqrt{\frac{11}{2}}t\right)y(t)=5cos(211​​t)+8112​​sin(211​​t).

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1. Solve the characteristic equation.
2. Write the general solution.
3. Apply the initial conditions.

Let's solve the characteristic equation!
We replace $y''$ with $r^2$ and $y$ with $1$ in the differential equation $6y'' + 33y = 0$ to get the characteristic equation $6r^2 + 33 = 0$ .

The solution to the differential equation is $y(t) = 5\cos\left(\sqrt{\frac{11}{2}}t\right) + 8\sqrt{\frac{2}{11}}\sin\left(\sqrt{\frac{11}{2}}t\right)$ .