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Magazine Chapter 9: Problem 15
Find the solution of the differential equation that satisfies the given initial condition. $$\frac{d u}{d t}=\frac{2 t+\sec ^{2} t}{2 u}, \quad u(0)=-5$$
Short Answer
Expert verified
The solution is \( u = -\sqrt{t^2 + \tan t + 25} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( \frac{d u}{d t} = \frac{2 t + \sec^2 t}{2 u} \). This equation is separable, meaning we can rewrite it in the form where all terms involving \( u \) are on one side and all terms involving \( t \) are on the other.
02
Separate the Variables
To separate the variables, multiply both sides by \( 2u \): \[ 2u \frac{d u}{d t} = 2t + \sec^2 t \]Then, rearrange to get each variable on its respective side:\[ 2u \, du = (2t + \sec^2 t) \, dt \]
03
Integrate Both Sides
Now integrate both sides of the equation:\[ \int 2u \, du = \int (2t + \sec^2 t) \, dt \]The left side integrates to \( u^2 + C_1 \). The right side integrates to \( t^2 + \tan t + C_2 \).
04
Simplify the Result
Combining the constants of integration \(C_1\) and \(C_2\) into a single constant \(C\), we have:\[ u^2 = t^2 + \tan t + C \]
05
Apply the Initial Condition
Substitute the initial condition \(u(0) = -5\) into the equation to find \(C\):\[ (-5)^2 = 0^2 + \tan(0) + C \]This simplifies to\[ 25 = 0 + 0 + C \]Therefore, \(C = 25\).
06
Write the General Solution with Constant
After applying the initial condition, the final expression is:\[ u^2 = t^2 + \tan t + 25 \]
07
Solve for u
Solving for \( u \), we take the square root:\[ u = \pm \sqrt{t^2 + \tan t + 25} \]Considering the initial condition \(u(0) = -5\), we find that\[ u = -\sqrt{t^2 + \tan t + 25} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Differential Equations
A separable differential equation is one in which you can split the different variables into separate sides of the equation. This makes it easier to solve because you can deal with each variable individually. In this process, you essentially "separate" the variables. This means re-arranging the equation so that all instances of one variable, such as \( u \), are on one side, while all instances of the other variable, such as \( t \), are on the other.
In the given problem, the differential equation \( \frac{d u}{d t} = \frac{2 t + \sec^2 t}{2 u} \) is separable. By multiplying both sides by \( 2u \), we get \( 2u \frac{d u}{d t} = 2t + \sec^2 t \), which can be rewritten to facilitate integration as \( 2u \, du = (2t + \sec^2 t) \, dt \). Through separation, we can now integrate each side with respect to its own variable.
In the given problem, the differential equation \( \frac{d u}{d t} = \frac{2 t + \sec^2 t}{2 u} \) is separable. By multiplying both sides by \( 2u \), we get \( 2u \frac{d u}{d t} = 2t + \sec^2 t \), which can be rewritten to facilitate integration as \( 2u \, du = (2t + \sec^2 t) \, dt \). Through separation, we can now integrate each side with respect to its own variable.
Initial Condition
An initial condition specifies the value of the solution at a particular point and is essential for finding a unique solution to a differential equation. In many problems, like the one we are discussing, the initial condition helps determine the constant of integration.
For the given exercise, the initial condition provided is \( u(0) = -5 \). This means that when the variable \( t \) is zero, the solution \( u \) should be \(-5\).
When you integrate a separable differential equation, you usually end up with an expression that has a constant \( C \). By using the initial condition, you can substitute \( t = 0 \) and \( u = -5 \) into your integrated result to solve for \( C \). This was done in Step 5 of the solution, where it led to finding that \( C = 25 \). That step ensures the solution of the equation is tailored to include the initial condition.
For the given exercise, the initial condition provided is \( u(0) = -5 \). This means that when the variable \( t \) is zero, the solution \( u \) should be \(-5\).
When you integrate a separable differential equation, you usually end up with an expression that has a constant \( C \). By using the initial condition, you can substitute \( t = 0 \) and \( u = -5 \) into your integrated result to solve for \( C \). This was done in Step 5 of the solution, where it led to finding that \( C = 25 \). That step ensures the solution of the equation is tailored to include the initial condition.
Integration
Integration is a key operation in solving differential equations. It is the mathematical process of finding the integral of a function and is the reverse operation of differentiation. When solving a separable differential equation, you integrate both sides to find the solution.
In Step 3 of the original solution, the equation \( 2u \, du = (2t + \sec^2 t) \, dt \) was integrated on both sides. Integrating \( 2u \, du \) yields \( u^2 + C_1 \), whereas integrating \( (2t + \sec^2 t) \, dt \) results in \( t^2 + \tan t + C_2 \).
Notice how every integral has a constant of integration鈥攗sually denoted by \( C \)鈥攖hat must be determined later on, usually using the initial condition. Integration thus transforms the differential form of a solution into a more interpretable expression that can be solved explicitly.
In Step 3 of the original solution, the equation \( 2u \, du = (2t + \sec^2 t) \, dt \) was integrated on both sides. Integrating \( 2u \, du \) yields \( u^2 + C_1 \), whereas integrating \( (2t + \sec^2 t) \, dt \) results in \( t^2 + \tan t + C_2 \).
Notice how every integral has a constant of integration鈥攗sually denoted by \( C \)鈥攖hat must be determined later on, usually using the initial condition. Integration thus transforms the differential form of a solution into a more interpretable expression that can be solved explicitly.
Solution of Differential Equations
Solving a differential equation means finding a function or set of functions that satisfy the equation. Once you have separated the variables and integrated both sides, the final step involves writing down or expressing the solution of the differential equation.
In Step 6, the integrated result \( u^2 = t^2 + \tan t + 25 \) is presented. However, to express \( u \) as a function of \( t \), solving it required taking the square root of both sides, which led to \( u = \pm \sqrt{t^2 + \tan t + 25} \).
Due to the initial condition \( u(0) = -5 \), we choose the negative branch of the square root to ensure \( u = -\sqrt{t^2 + \tan t + 25} \).
This results in a solution that fits both the differential equation and the given initial condition, completing the problem-solving process by finding a specific function that works throughout the problem's scope.
In Step 6, the integrated result \( u^2 = t^2 + \tan t + 25 \) is presented. However, to express \( u \) as a function of \( t \), solving it required taking the square root of both sides, which led to \( u = \pm \sqrt{t^2 + \tan t + 25} \).
Due to the initial condition \( u(0) = -5 \), we choose the negative branch of the square root to ensure \( u = -\sqrt{t^2 + \tan t + 25} \).
This results in a solution that fits both the differential equation and the given initial condition, completing the problem-solving process by finding a specific function that works throughout the problem's scope.
