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Chapter 7: Problem 3

Find the solution to the differential equation $$ \begin{array}{l} \frac{d y}{d t}=y^{2}(5+t), \\ y=5 \text { when } t=1 . \end{array} $$

Short Answer

Expert verified
The solution is \( y = -\frac{1}{\frac{5}{2} t^2 + 5 t - \frac{11}{2}}\).

Step by step solution

01

Identify the Type of Differential Equation

Recognize that the given equation \(\frac{d y}{d t}=y^{2}(5+t)\) is a separable differential equation because the variables \(y\) and \(t\) can be separated to different sides of the equation.
02

Separate the Variables

Rewrite the equation by separating the variables: \(\frac{1}{y^2} \frac{d y}{dt} = 5 + t\). This can be further simplified to \(\frac{1}{y^2} d y = (5 + t) dt\).
03

Integrate Both Sides

Integrate both sides to find \(\frac{-1}{y} = \frac{5}{2} t^2 + 5 t + C\). Integrate \(\frac{1}{y^2} d y\) to get \(-\frac{1}{y}\) and integrate \((5 + t) dt\) to get \(\frac{5}{2} t^2 + 5 t + C\).
04

Solve for the Constant of Integration

Use the initial condition \(y = 5\) when \(t = 1\) to solve for \(C\). Substitute \(y = 5\) and \(t = 1\) into \(\frac{-1}{y} = \frac{5}{2} t^2 + 5 t + C\) to get \(\frac{-1}{5} = \frac{5}{2} (1)^2 + 5(1) + C\). Simplifying this will give \(C = \frac{-11}{2}\).
05

Construct the General Solution

Substitute \(C\) back into the integrated equation to get \(\frac{-1}{y} = \frac{5}{2} t^2 + 5 t - \frac{11}{2}\).
06

Solve for \(y\)

Finally, solve for \(y\) by taking the reciprocal of both sides and simplifying. \(y = -\frac{1}{\frac{5}{2} t^2 + 5 t - \frac{11}{2}}\).

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Key Concepts

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

Separable Differential Equations
In differential equations, a separable differential equation is one that can be written in the form \(\frac{dy}{dt} = g(y) h(t)\). This means that the differential equation can be separated into two parts: one involving only the dependent variable \(y\) and its differential \(dy\), and the other involving only the independent variable \(t\) and its differential \(dt\).
Initial Conditions in Differential Equations
When solving differential equations, initial conditions are very important. They are used to find the exact solution to a differential equation. Initial conditions provide specific values for the variables at a particular point. This helps to determine the constants of integration that appear after integrating the differential equation.

In the provided exercise, the initial condition is given as \(y = 5\) when \(t = 1\). This means that at time \(t = 1\), the value of \(y\) is 5. These conditions help to find the constant \(C\) after integrating the equation.
Integration of Differential Equations
To solve a separable differential equation, we integrate both sides of the separated equation. After separating the variables, integrate both sides to find the solution.

For example, the equation \(\frac{1}{y^2} dy = (5 + t) dt\) was integrated to obtain:
  • \( -\frac{1}{y} = \frac{5}{2}t^2 + 5t + C\)
.

Here:
  • The left side \(-\frac{1}{y}\) is the result of integrating \(\frac{1}{y^2} dy\).
  • The right side is the result of integrating \((5 + t) dt\), which gives \( \frac{5}{2} t^2 + 5t + C \).
.

Using the initial condition to determine \( C\), the final step is to plug this value back into the equation, which allows us to solve for \(y\). This provides the complete and specific solution to the differential equation.

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Most popular questions from this chapter

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

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

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?

Find the equation of the solution to \(\frac{d y}{d x}=x^{2} y\) through the point \((x, y)=(1,3)\). help (equations)

Suppose that a cylindrical water tank with a hole in the bottom is filled with water. The water, of course, will leak out and the height of the water will decrease. Let \(h(t)\) denote the height of the water. A physical principle called Torricelli's Law implies that the height decreases at a rate proportional to the square root of the height. a. Express this fact using \(k\) as the constant of proportionality. b. Suppose you have two tanks, one with \(k=-1\) and another with \(k=-10 .\) What physical differences would you expect to find? c. Suppose you have a tank for which the height decreases at 20 inches per minute when the water is filled to a depth of 100 inches. Find the value of \(k\). d. Solve the initial value problem for the tank in part (c), and graph the solution you determine. e. How long does it take for the water to run out of the tank? f. Is the solution that you found valid for all time \(t ?\) If so, explain how you know this. If not, explain why not.
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