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Chapter 7: Problem 5

Find an equation of the curve that satisfies $$ \frac{d y}{d x}=45 y x^{4} $$ and whose \(y\) -intercept is \(5 .\) \(y(x)=\)

Short Answer

Expert verified
The equation of the curve is \(y = 5 e^{9x^5}\).

Step by step solution

01

Write Down the Differential Equation

The given differential equation is: \(\frac{dy}{dx} = 45yx^4\).
02

Separate the Variables

To solve the differential equation, separate the variables by dividing both sides by \(y\) and multiplying both sides by \(dx\): \(\frac{1}{y} dy = 45x^4 dx\).
03

Integrate Both Sides

Integrate both sides to find the general solution: \(\int\frac{1}{y} dy = \int 45x^4 dx\). This gives: \(\text{ln}|y| = \frac{45}{5}x^5 + C\). Simplifying, we get: \(\text{ln}|y| = 9x^5 + C\).
04

Exponentiate to Remove the Natural Log

Exponentiate both sides to solve for \(y\): \(e^{\text{ln}|y|} = e^{9x^5 + C}\). This simplifies to \(|y| = e^{9x^5} e^C\). Let \(e^C = C'\), then \(y = C' e^{9x^5}\).
05

Determine the Constant Using the Initial Condition

Use the given initial condition \(y(0) = 5\) to find the constant \(C'\). Substituting \(x = 0\) and \(y = 5\) into the equation: \(5 = C' e^{0}\). Therefore, \(C' = 5\).
06

Write the Final Solution

Substitute \(C' = 5\) back into the equation: \(y = 5 e^{9x^5}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
To solve the differential equation \(\frac{dy}{dx} = 45yx^4\), we use a method called separation of variables. This technique allows us to separate the terms involving \(y\) and \(x\) on opposite sides of the equation.
First, we rewrite the equation to isolate \(dy\) and \(dx\). By dividing both sides by \(y\) and multiplying both sides by \(dx\), we get:
\( \frac{1}{y} dy = 45x^4 dx \).
Now, the terms involving \(y\) are on one side and those involving \(x\) are on the other, making it possible to integrate each side independently.
Integration
The next step after separating the variables is integration. We integrate both sides of the separated equation to find the general solution.
Integrating \(\frac{1}{y} dy\) with respect to \(y\) and \(45x^4 dx\) with respect to \(x\), we get:
\(\text{ln}|y| = \frac{45}{5}x^5 + C\).
We simplify this to:
\(\text{ln}|y| = 9x^5 + C\)
where \(C\) is the constant of integration.
To solve for \(y\), we exponentiate both sides to remove the natural logarithm, leading to:
\(|y| = e^{9x^5 + C}\).
By letting \(e^C = C'\), a new constant, we find:
\(y = C' e^{9x^5}\).
Initial Conditions
Initial conditions help determine the specific solution of a differential equation. In our example, the initial condition given is \(y(0) = 5\).
We use this to find the constant \(C'\) in our general solution \(y = C' e^{9x^5}\).
Substituting \(x = 0\) and \(y = 5\), we get:
\(5 = C' e^{0}\).
This simplifies to \(C' = 5\).
Finally, substituting back into the general solution, we obtain the specific solution:
\(y = 5 e^{9x^5}\).
Initial conditions are a crucial part of solving differential equations as they enable us to find a unique solution that fits the given context.

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Most popular questions from this chapter

A bacteria culture starts with 220 bacteria and grows at a rate proportional to its size. After 5 hours there will be 1100 bacteria. (a) Express the population after \(t\) hours as a function of \(t\). population: (function of (b) What will be the population after 5 hours? (c) How long will it take for the population to reach \(1510 ?\)

In this problem, we test further what it means for a function to be a solution to a given differential equation. a. Consider the differential equation $$ \frac{d y}{d t}=y-t $$ Determine whether the following functions are solutions to the given differential equation. $$ \begin{array}{l} \text { - } y(t)=t+1+2 e^{t} \\ \text { - } y(t)=t+1 \\ \text { - } y(t)=t+2 \end{array} $$ b. When you weigh bananas in a scale at the grocery store, the height \(h\) of the bananas is described by the differential equation $$ \frac{d^{2} h}{d t^{2}}=-k h $$ where \(k\) is the spring constant, a constant that depends on the properties of the spring in the scale. After you put the bananas in the scale, you (cleverly) observe that the height of the bananas is given by \(h(t)=4 \sin (3 t) .\) What is the value of the spring constant?

Congratulations, you just won the lottery! In one option presented to you, you will be paid one million dollars a year for the next 25 years. You can deposit this money in an account that will earn \(5 \%\) each year. a. Set up a differential equation that describes the rate of change in the amount of money in the account. Two factors cause the amount to grow-first, you are depositing one millon dollars per year and second, you are earning \(5 \%\) interest. b. If there is no amount of money in the account when you open it, how much money will you have in the account after 25 years? c. The second option presented to you is to take a lump sum of 10 million dollars, which you will deposit into a similar account. How much money will you have in that account after 25 years? d. Do you prefer the first or second option? Explain your thinking. e. At what time does the amount of money in the account under the first option overtake the amount of money in the account under the second option?

Consider the solution of the differential equation \(y^{\prime}=-3 y\) passing through \(y(0)=1.5\). A. Sketch the slope field for this differential equation, and sketch the solution passing through the point (0,1.5) B. Use Euler's method with step size \(\Delta x=0.2\) to estimate the solution at \(x=0.2,0.4, \ldots, 1\), using these to fill in the following table. (Be sure not to round your answers at each step!) \begin{tabular}{|l|l|l|l|l|l|l|} \hline\(x=\) & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline\(y \approx\) & 1.5 & & & & & \\ \hline \end{tabular} C. Plot your estimated solution on your slope field. Compare the solution and the slope field. Is the estimated solution an over or under estimate for the actual solution? D. Check that \(y=1.5 e^{-3 x}\) is a solution to \(y^{\prime}=-3 y\) with \(y(0)=1.5\).

Solve the separable differential equation for \(u\) $$ \frac{d u}{d t}=e^{2 u+8 t} $$ Use the following initial condition: \(u(0)=13\). \(=\)
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