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Chapter 10: Problem 7

Solve the following differential equations: $$ y^{\prime}=\left(\frac{t}{y}\right)^{2} e^{t^{3}} $$

Short Answer

Expert verified
The general solution is \(y = \sqrt[3]{e^{t^3} + 3C}\)

Step by step solution

01

Separate the variables

Rewrite the given differential equation to separate the variables, placing all terms involving \(y\) on one side and all terms involving \(t\) on the other side. Starting with \(y' = \frac{t^2 e^{t^3}}{y^2}\), multiply both sides by \(y^2 dt\) to get \( y^2 dy = t^2 e^{t^3} dt\).
02

Integrate both sides

Integrate both sides of the equation:\begin{align*} \int y^2 dy &= \int t^2 e^{t^3} dt \end{align*}The left side integrates to \( \int y^2 dy= \frac{y^3}{3} + C_1\).For the right side, let \(u = t^3\), then \(du = 3t^2 dt\), substitute and integrate: \( \int t^2 e^{t^3} dt= \frac{1}{3} \int e^u du = \frac{1}{3} e^u + C_2= \frac{1}{3}e^{t^3} + C_2\).
03

Simplify and combine constants

Combine the constants of integration \(C_1\) and \(C_2\) into a single constant: \( \frac{y^3}{3} = \frac{1}{3} e^{t^3} + C\)Multiply through by 3 to simplify: \( y^3 = e^{t^3} + 3C\).
04

Solve for \(y\)

Solve for \(y\) to get the general solution: \( y = \sqrt[3]{e^{t^3} + 3C}\)

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Key Concepts

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

Variable Separation
Variable separation is a fundamental technique used to solve differential equations. The idea is to rearrange the equation so that all terms involving one variable (e.g., y) are on one side, while all terms involving the other variable (e.g., t) are on the opposite side. This enables us to integrate both sides separately.

In our example, we start with\( y' = \frac{t^2 e^{t^3}}{y^2}\). To separate the variables, multiply both sides by\( y^2 dt\): \(y^2 dy = t^2 e^{t^3} dt\). Now, all the y terms are on the left, and all the t terms are on the right. This allows us to proceed to the next step, integration.
Integration
Once the variables are separated, we integrate both sides of the equation. Integration is the process of finding the antiderivative of a function. Here, we integrate \(\int y^2 dy \) on the left and \(\int t^2 e^{t^3} dt \) on the right.

For the left side, the integral is straightforward: \(\int y^2 dy = \frac{y^3}{3} + C_1\). For the right side, things are more complex and we might need a substitution. Therefore, we move to the next useful technique: integration by substitution.
Integration by Substitution
Integration by substitution is used to simplify the integral of a complex function. The goal is to replace a complicated part of the function with a new variable, making the integral easier.

For the right side of our equation \(\int t^2 e^{t^3} dt\), we set \(u = t^3\). Then, differentiate to find \(du = 3t^2 dt\) or \(t^2 dt = \frac{1}{3} du\). We substitute these into the integral: \( \int t^2 e^{t^3} dt = \frac{1}{3} \int e^u du \). This simplifies to \( \frac{1}{3} e^u + C_2 = \frac{1}{3} e^{t^3} + C_2 \).
With this approach, we have successfully integrated both sides.
General Solution
The general solution is the expression involving all possible solutions of the differential equation. After integration, we have two constants of integration, \(C_1\) and \(C_2\). To combine them, we write \( \frac{y^3}{3} = \frac{1}{3} e^{t^3} + C\), where \(C = C_2 - C_1\).

Multiplying through by 3, we obtain: \( y^3 = e^{t^3} + 3C\). Finally, solving for \(y\), we get the general solution: \( y = \sqrt[3]{e^{t^3} + 3C} \). This final expression includes the arbitrary constant C, representing all possible solutions.

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Most popular questions from this chapter

Suppose that \(f(t)\) is a solution of \(y^{\prime}=t^{2}-y^{2}\) and the graph of \(f(t)\) passes through the point \((2,3)\). Find the slope of the graph when \(t=2\).

Let \(f(t)\) be the solution of \(y^{\prime}=-(t+1) y^{2}, y(0)=1\). Use Euler's method with \(n=5\) to estimate \(f(1) .\) Then, solve the differential equation, find an explicit formula for \(f(t)\) and compute \(f(1)\). How accurate is the estimated value of \(f(1)\) ?

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 5 ;(0,3)\) is on the graph; the slope is always negative, and the slope becomes less negative.

One problem in psychology is to determine the relation between some physical stimulus and the corresponding sensation or reaction produced in a subject. Suppose that, measured in appropriate units, the strength of a stimulus is \(s\) and the intensity of the corresponding sensation is some function of \(s\), say, \(f(s)\). Some experimental data suggest that the rate of change of intensity of the sensation with respect to the stimulus is directly proportional to the intensity of the sensation and inversely proportional to the strength of the stimulus; that is, \(f(s)\) satisfies the differential equation $$ \frac{d y}{d s}=k \frac{y}{s} $$ for some positive constant \(k .\) Solve this differential equation. (Figure 7 shows several solutions corresponding to \(k=.4 .)\)

Solve the following differential equations with the given initial conditions. $$ y^{\prime}=\frac{t^{2}}{y}, y(0)=-5 $$
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