Solved on Feb 27, 2024

Find the limit of the solution to the differential equation $y' = y^3 - 4y$ with initial condition $y(0) = 0.2$ as $t$ approaches infinity.

STEP 1

Assumptions
1. The given differential equation is $y^{\prime} = y^{3} - 4y$ .
2. The initial condition is $y(0) = 0.2$ .
3. We are interested in the behavior of the solution as $t$ approaches infinity.

STEP 2

To solve the differential equation, we can use separation of variables. We want to separate the $y$ terms and the $t$ terms on opposite sides of the equation.
$\frac{dy}{dt} = y^{3} - 4y$

STEP 3

Separate the variables by dividing both sides by $y^{3} - 4y$ and multiplying both sides by $dt$ .
$\frac{1}{y^{3} - 4y}dy = dt$

STEP 4

To integrate the left side, we need to factor the denominator.
$y^{3} - 4y = y(y^{2} - 4) = y(y - 2)(y + 2)$

STEP 5

Now we can write the partial fraction decomposition.
$\frac{1}{y(y - 2)(y + 2)} = \frac{A}{y} + \frac{B}{y - 2} + \frac{C}{y + 2}$

STEP 6

To find the constants $A$ , $B$ , and $C$ , we multiply both sides by the denominator $y(y - 2)(y + 2)$ and then equate coefficients or use values that simplify the equation.
$1 = A(y - 2)(y + 2) + B(y)(y + 2) + C(y)(y - 2)$

STEP 7

Let's find $A$ by setting $y = 0$ .
$1 = A(-2)(2)$
$A = -\frac{1}{4}$

STEP 8

Find $B$ by setting $y = 2$ .
$1 = B(2)(4)$
$B = \frac{1}{8}$

STEP 9

Find $C$ by setting $y = -2$ .
$1 = C(-2)(-4)$
$C = \frac{1}{8}$

STEP 10

Now that we have the constants $A$ , $B$ , and $C$ , we can write the integral.
$\int \frac{1}{y(y - 2)(y + 2)}dy = \int \left(-\frac{1}{4y} + \frac{1}{8(y - 2)} + \frac{1}{8(y + 2)}\right) dy$

STEP 11

Integrate both sides with respect to $y$ .
$\int \left(-\frac{1}{4y} + \frac{1}{8(y - 2)} + \frac{1}{8(y + 2)}\right) dy = -\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| + C_1$

STEP 12

Integrate the right side with respect to $t$ .
$\int dt = t + C_2$

STEP 13

Combine the constants of integration into a single constant $C$ .
$C = C_1 - C_2$

STEP 14

Now we have the general solution of the differential equation.
$-\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| = t + C$

STEP 15

Apply the initial condition $y(0) = 0.2$ to find the constant $C$ .
$-\frac{1}{4}\ln|0.2| + \frac{1}{8}\ln|0.2 - 2| + \frac{1}{8}\ln|0.2 + 2| = 0 + C$

STEP 16

Calculate the constant $C$ .
$C = -\frac{1}{4}\ln|0.2| + \frac{1}{8}\ln|-1.8| + \frac{1}{8}\ln|2.2|$

STEP 17

Simplify the expression for $C$ by calculating the logarithms.
$C = -\frac{1}{4}\ln(0.2) + \frac{1}{8}\ln(1.8) + \frac{1}{8}\ln(2.2)$

STEP 18

Now we have the specific solution that satisfies the initial condition.
$-\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| = t + C$

STEP 19

To find the limit as $t$ approaches infinity, we need to analyze the behavior of the solution. Since the solution is an increasing function of $t$ (as can be seen from the differential equation $y^{\prime} = y^{3} - 4y$ ), we expect $y(t)$ to approach a stable equilibrium point as $t$ goes to infinity.

STEP 20

The equilibrium points of the differential equation are the values of $y$ that make $y^{\prime} = 0$ .
$y^{3} - 4y = 0$

STEP 21

Solve for the equilibrium points.
$y(y^2 - 4) = 0$
$y(y - 2)(y + 2) = 0$

STEP 22

The equilibrium points are $y = 0$ , $y = 2$ , and $y = -2$ .

STEP 23

Since the initial condition is $y(0) = 0.2$ , which is between $0$ and $2$ , and the solution is increasing, we expect the solution to approach the next equilibrium point, which is $y = 2$ .

STEP 24

Therefore, the limit of $y(t)$ as $t$ approaches infinity is the stable equilibrium point $y = 2$ .
$\lim_{t \rightarrow \infty} y(t) = 2$
The solution to the limit is $2$ .

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Find the limit of the solution to the differential equation y′=y3−4yy' = y^3 - 4yy′=y3−4y with initial condition y(0)=0.2y(0) = 0.2y(0)=0.2 as ttt approaches infinity.

STEP 1

Assumptions1. The given differential equation is y′=y3−4yy^{\prime} = y^{3} - 4yy′=y3−4y.2. The initial condition is y(0)=0.2y(0) = 0.2y(0)=0.2.3. We are interested in the behavior of the solution as ttt approaches infinity.

STEP 2

To solve the differential equation, we can use separation of variables. We want to separate the yyy terms and the ttt terms on opposite sides of the equation.dydt=y3−4y\frac{dy}{dt} = y^{3} - 4ydtdy​=y3−4y

STEP 3

Separate the variables by dividing both sides by y3−4yy^{3} - 4yy3−4y and multiplying both sides by dtdtdt.1y3−4ydy=dt\frac{1}{y^{3} - 4y}dy = dty3−4y1​dy=dt

STEP 4

To integrate the left side, we need to factor the denominator.y3−4y=y(y2−4)=y(y−2)(y+2)y^{3} - 4y = y(y^{2} - 4) = y(y - 2)(y + 2)y3−4y=y(y2−4)=y(y−2)(y+2)

STEP 5

Now we can write the partial fraction decomposition.1y(y−2)(y+2)=Ay+By−2+Cy+2\frac{1}{y(y - 2)(y + 2)} = \frac{A}{y} + \frac{B}{y - 2} + \frac{C}{y + 2}y(y−2)(y+2)1​=yA​+y−2B​+y+2C​

STEP 6

STEP 7

Let's find AAA by setting y=0y = 0y=0.1=A(−2)(2)1 = A(-2)(2)1=A(−2)(2)A=−14A = -\frac{1}{4}A=−41​

STEP 8

Find BBB by setting y=2y = 2y=2.1=B(2)(4)1 = B(2)(4)1=B(2)(4)B=18B = \frac{1}{8}B=81​

STEP 9

Find CCC by setting y=−2y = -2y=−2.1=C(−2)(−4)1 = C(-2)(-4)1=C(−2)(−4)C=18C = \frac{1}{8}C=81​

STEP 10

STEP 11

STEP 12

Integrate the right side with respect to ttt.∫dt=t+C2\int dt = t + C_2∫dt=t+C2​

STEP 13

Combine the constants of integration into a single constant CCC.C=C1−C2C = C_1 - C_2C=C1​−C2​

STEP 14

Now we have the general solution of the differential equation.−14ln⁡∣y∣+18ln⁡∣y−2∣+18ln⁡∣y+2∣=t+C-\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| = t + C−41​ln∣y∣+81​ln∣y−2∣+81​ln∣y+2∣=t+C

STEP 15

Apply the initial condition y(0)=0.2y(0) = 0.2y(0)=0.2 to find the constant CCC.−14ln⁡∣0.2∣+18ln⁡∣0.2−2∣+18ln⁡∣0.2+2∣=0+C-\frac{1}{4}\ln|0.2| + \frac{1}{8}\ln|0.2 - 2| + \frac{1}{8}\ln|0.2 + 2| = 0 + C−41​ln∣0.2∣+81​ln∣0.2−2∣+81​ln∣0.2+2∣=0+C

STEP 16

Calculate the constant CCC.C=−14ln⁡∣0.2∣+18ln⁡∣−1.8∣+18ln⁡∣2.2∣C = -\frac{1}{4}\ln|0.2| + \frac{1}{8}\ln|-1.8| + \frac{1}{8}\ln|2.2|C=−41​ln∣0.2∣+81​ln∣−1.8∣+81​ln∣2.2∣

STEP 17

Simplify the expression for CCC by calculating the logarithms.C=−14ln⁡(0.2)+18ln⁡(1.8)+18ln⁡(2.2)C = -\frac{1}{4}\ln(0.2) + \frac{1}{8}\ln(1.8) + \frac{1}{8}\ln(2.2)C=−41​ln(0.2)+81​ln(1.8)+81​ln(2.2)

STEP 18

Now we have the specific solution that satisfies the initial condition.−14ln⁡∣y∣+18ln⁡∣y−2∣+18ln⁡∣y+2∣=t+C-\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| = t + C−41​ln∣y∣+81​ln∣y−2∣+81​ln∣y+2∣=t+C

STEP 19

STEP 20

The equilibrium points of the differential equation are the values of yyy that make y′=0y^{\prime} = 0y′=0.y3−4y=0y^{3} - 4y = 0y3−4y=0

STEP 21

Solve for the equilibrium points.y(y2−4)=0y(y^2 - 4) = 0y(y2−4)=0y(y−2)(y+2)=0y(y - 2)(y + 2) = 0y(y−2)(y+2)=0

STEP 22

The equilibrium points are y=0y = 0y=0, y=2y = 2y=2, and y=−2y = -2y=−2.

STEP 23

Since the initial condition is y(0)=0.2y(0) = 0.2y(0)=0.2, which is between 000 and 222, and the solution is increasing, we expect the solution to approach the next equilibrium point, which is y=2y = 2y=2.

STEP 24

Therefore, the limit of y(t)y(t)y(t) as ttt approaches infinity is the stable equilibrium point y=2y = 2y=2.lim⁡t→∞y(t)=2\lim_{t \rightarrow \infty} y(t) = 2t→∞lim​y(t)=2The solution to the limit is 222.

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Find the limit of the solution to the differential equation $y' = y^3 - 4y$ with initial condition $y(0) = 0.2$ as $t$ approaches infinity.

Assumptions
1. The given differential equation is $y^{\prime} = y^{3} - 4y$ .
2. The initial condition is $y(0) = 0.2$ .
3. We are interested in the behavior of the solution as $t$ approaches infinity.

To solve the differential equation, we can use separation of variables. We want to separate the $y$ terms and the $t$ terms on opposite sides of the equation.
$\frac{dy}{dt} = y^{3} - 4y$

Separate the variables by dividing both sides by $y^{3} - 4y$ and multiplying both sides by $dt$ .
$\frac{1}{y^{3} - 4y}dy = dt$

To integrate the left side, we need to factor the denominator.
$y^{3} - 4y = y(y^{2} - 4) = y(y - 2)(y + 2)$

Now we can write the partial fraction decomposition.
$\frac{1}{y(y - 2)(y + 2)} = \frac{A}{y} + \frac{B}{y - 2} + \frac{C}{y + 2}$

Let's find $A$ by setting $y = 0$ .
$1 = A(-2)(2)$
$A = -\frac{1}{4}$

Find $B$ by setting $y = 2$ .
$1 = B(2)(4)$
$B = \frac{1}{8}$

Find $C$ by setting $y = -2$ .
$1 = C(-2)(-4)$
$C = \frac{1}{8}$

Integrate the right side with respect to $t$ .
$\int dt = t + C_2$

Combine the constants of integration into a single constant $C$ .
$C = C_1 - C_2$

Now we have the general solution of the differential equation.
$-\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| = t + C$

Apply the initial condition $y(0) = 0.2$ to find the constant $C$ .
$-\frac{1}{4}\ln|0.2| + \frac{1}{8}\ln|0.2 - 2| + \frac{1}{8}\ln|0.2 + 2| = 0 + C$

Calculate the constant $C$ .
$C = -\frac{1}{4}\ln|0.2| + \frac{1}{8}\ln|-1.8| + \frac{1}{8}\ln|2.2|$

Simplify the expression for $C$ by calculating the logarithms.
$C = -\frac{1}{4}\ln(0.2) + \frac{1}{8}\ln(1.8) + \frac{1}{8}\ln(2.2)$

Now we have the specific solution that satisfies the initial condition.
$-\frac{1}{4}\ln|y| + \frac{1}{8}\ln|y - 2| + \frac{1}{8}\ln|y + 2| = t + C$

The equilibrium points of the differential equation are the values of $y$ that make $y^{\prime} = 0$ .
$y^{3} - 4y = 0$

Solve for the equilibrium points.
$y(y^2 - 4) = 0$
$y(y - 2)(y + 2) = 0$

The equilibrium points are $y = 0$ , $y = 2$ , and $y = -2$ .

Since the initial condition is $y(0) = 0.2$ , which is between $0$ and $2$ , and the solution is increasing, we expect the solution to approach the next equilibrium point, which is $y = 2$ .

Therefore, the limit of $y(t)$ as $t$ approaches infinity is the stable equilibrium point $y = 2$ .
$\lim_{t \rightarrow \infty} y(t) = 2$
The solution to the limit is $2$ .