Solved on Feb 22, 2024

Find the general solution of the second-order linear ODE $y'' + y = 0$ .

STEP 1

Assumptions
1. We are given a second-order linear ordinary differential equation (ODE) in the form $y^{\prime \prime} + y = 0$ .
2. The solution to this ODE will involve functions whose second derivatives are related to the function itself.
3. We will use the characteristic equation method to find the solution.

STEP 2

To solve the second-order linear ODE, we will look for solutions of the form $y = e^{rt}$ , where $r$ is a constant to be determined and $t$ is the independent variable.

STEP 3

Substitute $y = e^{rt}$ into the ODE to find the characteristic equation. The first derivative $y^{\prime}$ is $re^{rt}$ and the second derivative $y^{\prime \prime}$ is $r^2e^{rt}$ .
$y^{\prime \prime} + y = r^2e^{rt} + e^{rt} = 0$

STEP 4

Factor out $e^{rt}$ from the equation since $e^{rt}$ is never zero.
$e^{rt}(r^2 + 1) = 0$

STEP 5

Set the expression in parentheses equal to zero to find the characteristic equation.
$r^2 + 1 = 0$

STEP 6

Solve the characteristic equation for $r$ .
$r^2 = -1$

STEP 7

Take the square root of both sides of the equation to find $r$ .
$r = \pm \sqrt{-1}$

STEP 8

Since $\sqrt{-1}$ is the imaginary unit $i$ , we have two roots: $r_1 = i$ and $r_2 = -i$ .

STEP 9

The general solution to the ODE is a linear combination of the solutions corresponding to $r_1$ and $r_2$ .
$y(t) = C_1e^{r_1t} + C_2e^{r_2t}$

STEP 10

Substitute $r_1 = i$ and $r_2 = -i$ into the general solution.
$y(t) = C_1e^{it} + C_2e^{-it}$

STEP 11

Euler's formula states that $e^{ix} = \cos(x) + i\sin(x)$ . Use Euler's formula to express $e^{it}$ and $e^{-it}$ .
$e^{it} = \cos(t) + i\sin(t)$ $e^{-it} = \cos(-t) + i\sin(-t)$

STEP 12

Since cosine is an even function and sine is an odd function, we have $\cos(-t) = \cos(t)$ and $\sin(-t) = -\sin(t)$ .
$e^{-it} = \cos(t) - i\sin(t)$

STEP 13

Substitute the expressions from Euler's formula into the general solution.
$y(t) = C_1(\cos(t) + i\sin(t)) + C_2(\cos(t) - i\sin(t))$

STEP 14

Combine like terms in the general solution.
$y(t) = (C_1 + C_2)\cos(t) + (C_1i - C_2i)\sin(t)$

STEP 15

Let $A = C_1 + C_2$ and $B = C_1i - C_2i$ . Since $C_1$ and $C_2$ are arbitrary constants, $A$ and $B$ are also arbitrary constants, and $B$ can be written as $Bi$ where $B$ is real.
$y(t) = A\cos(t) + Bi\sin(t)$

STEP 16

The final form of the solution to the ODE $y^{\prime \prime} + y = 0$ is:
$y(t) = A\cos(t) + B\sin(t)$
where $A$ and $B$ are arbitrary real constants determined by initial conditions.

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Find the general solution of the second-order linear ODE $y'' + y = 0$ .

STEP 1

STEP 2

To solve the second-order linear ODE, we will look for solutions of the form $y = e^{rt}$ , where $r$ is a constant to be determined and $t$ is the independent variable.

STEP 3

Substitute $y = e^{rt}$ into the ODE to find the characteristic equation. The first derivative $y^{\prime}$ is $re^{rt}$ and the second derivative $y^{\prime \prime}$ is $r^2e^{rt}$ .
$y^{\prime \prime} + y = r^2e^{rt} + e^{rt} = 0$

STEP 4

Factor out $e^{rt}$ from the equation since $e^{rt}$ is never zero.
$e^{rt}(r^2 + 1) = 0$

STEP 5

Set the expression in parentheses equal to zero to find the characteristic equation.
$r^2 + 1 = 0$

STEP 6

Solve the characteristic equation for $r$ .
$r^2 = -1$

STEP 7

Take the square root of both sides of the equation to find $r$ .
$r = \pm \sqrt{-1}$

STEP 8

Since $\sqrt{-1}$ is the imaginary unit $i$ , we have two roots: $r_1 = i$ and $r_2 = -i$ .

STEP 9

The general solution to the ODE is a linear combination of the solutions corresponding to $r_1$ and $r_2$ .
$y(t) = C_1e^{r_1t} + C_2e^{r_2t}$

STEP 10

Substitute $r_1 = i$ and $r_2 = -i$ into the general solution.
$y(t) = C_1e^{it} + C_2e^{-it}$

STEP 11

Euler's formula states that $e^{ix} = \cos(x) + i\sin(x)$ . Use Euler's formula to express $e^{it}$ and $e^{-it}$ .
$e^{it} = \cos(t) + i\sin(t)$ $e^{-it} = \cos(-t) + i\sin(-t)$

STEP 12

Since cosine is an even function and sine is an odd function, we have $\cos(-t) = \cos(t)$ and $\sin(-t) = -\sin(t)$ .
$e^{-it} = \cos(t) - i\sin(t)$

STEP 13

Substitute the expressions from Euler's formula into the general solution.
$y(t) = C_1(\cos(t) + i\sin(t)) + C_2(\cos(t) - i\sin(t))$

STEP 14

Combine like terms in the general solution.
$y(t) = (C_1 + C_2)\cos(t) + (C_1i - C_2i)\sin(t)$

STEP 15

Let $A = C_1 + C_2$ and $B = C_1i - C_2i$ . Since $C_1$ and $C_2$ are arbitrary constants, $A$ and $B$ are also arbitrary constants, and $B$ can be written as $Bi$ where $B$ is real.
$y(t) = A\cos(t) + Bi\sin(t)$

STEP 16

The final form of the solution to the ODE $y^{\prime \prime} + y = 0$ is:
$y(t) = A\cos(t) + B\sin(t)$
where $A$ and $B$ are arbitrary real constants determined by initial conditions.

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Find the general solution of the second-order linear ODE y′′+y=0y'' + y = 0y′′+y=0.

STEP 1

STEP 2

To solve the second-order linear ODE, we will look for solutions of the form y=erty = e^{rt}y=ert, where rrr is a constant to be determined and ttt is the independent variable.

STEP 3

STEP 4

Factor out erte^{rt}ert from the equation since erte^{rt}ert is never zero.ert(r2+1)=0e^{rt}(r^2 + 1) = 0ert(r2+1)=0

STEP 5

Set the expression in parentheses equal to zero to find the characteristic equation.r2+1=0r^2 + 1 = 0r2+1=0

STEP 6

Solve the characteristic equation for rrr.r2=−1r^2 = -1r2=−1

STEP 7

Take the square root of both sides of the equation to find rrr.r=±−1r = \pm \sqrt{-1}r=±−1​

STEP 8

Since −1\sqrt{-1}−1​ is the imaginary unit iii, we have two roots: r1=ir_1 = ir1​=i and r2=−ir_2 = -ir2​=−i.

STEP 9

The general solution to the ODE is a linear combination of the solutions corresponding to r1r_1r1​ and r2r_2r2​.y(t)=C1er1t+C2er2ty(t) = C_1e^{r_1t} + C_2e^{r_2t}y(t)=C1​er1​t+C2​er2​t

STEP 10

Substitute r1=ir_1 = ir1​=i and r2=−ir_2 = -ir2​=−i into the general solution.y(t)=C1eit+C2e−ity(t) = C_1e^{it} + C_2e^{-it}y(t)=C1​eit+C2​e−it

STEP 11

STEP 12

Since cosine is an even function and sine is an odd function, we have cos⁡(−t)=cos⁡(t)\cos(-t) = \cos(t)cos(−t)=cos(t) and sin⁡(−t)=−sin⁡(t)\sin(-t) = -\sin(t)sin(−t)=−sin(t).e−it=cos⁡(t)−isin⁡(t)e^{-it} = \cos(t) - i\sin(t)e−it=cos(t)−isin(t)

STEP 13

Substitute the expressions from Euler's formula into the general solution.y(t)=C1(cos⁡(t)+isin⁡(t))+C2(cos⁡(t)−isin⁡(t))y(t) = C_1(\cos(t) + i\sin(t)) + C_2(\cos(t) - i\sin(t))y(t)=C1​(cos(t)+isin(t))+C2​(cos(t)−isin(t))

STEP 14

Combine like terms in the general solution.y(t)=(C1+C2)cos⁡(t)+(C1i−C2i)sin⁡(t)y(t) = (C_1 + C_2)\cos(t) + (C_1i - C_2i)\sin(t)y(t)=(C1​+C2​)cos(t)+(C1​i−C2​i)sin(t)

STEP 15

STEP 16

The final form of the solution to the ODE y′′+y=0y^{\prime \prime} + y = 0y′′+y=0 is:y(t)=Acos⁡(t)+Bsin⁡(t)y(t) = A\cos(t) + B\sin(t)y(t)=Acos(t)+Bsin(t)where AAA and BBB are arbitrary real constants determined by initial conditions.

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Find the general solution of the second-order linear ODE $y'' + y = 0$ .

To solve the second-order linear ODE, we will look for solutions of the form $y = e^{rt}$ , where $r$ is a constant to be determined and $t$ is the independent variable.

Factor out $e^{rt}$ from the equation since $e^{rt}$ is never zero.
$e^{rt}(r^2 + 1) = 0$

Set the expression in parentheses equal to zero to find the characteristic equation.
$r^2 + 1 = 0$

Solve the characteristic equation for $r$ .
$r^2 = -1$

Take the square root of both sides of the equation to find $r$ .
$r = \pm \sqrt{-1}$

Since $\sqrt{-1}$ is the imaginary unit $i$ , we have two roots: $r_1 = i$ and $r_2 = -i$ .

The general solution to the ODE is a linear combination of the solutions corresponding to $r_1$ and $r_2$ .
$y(t) = C_1e^{r_1t} + C_2e^{r_2t}$

Substitute $r_1 = i$ and $r_2 = -i$ into the general solution.
$y(t) = C_1e^{it} + C_2e^{-it}$

Since cosine is an even function and sine is an odd function, we have $\cos(-t) = \cos(t)$ and $\sin(-t) = -\sin(t)$ .
$e^{-it} = \cos(t) - i\sin(t)$

Substitute the expressions from Euler's formula into the general solution.
$y(t) = C_1(\cos(t) + i\sin(t)) + C_2(\cos(t) - i\sin(t))$

Combine like terms in the general solution.
$y(t) = (C_1 + C_2)\cos(t) + (C_1i - C_2i)\sin(t)$

The final form of the solution to the ODE $y^{\prime \prime} + y = 0$ is:
$y(t) = A\cos(t) + B\sin(t)$
where $A$ and $B$ are arbitrary real constants determined by initial conditions.