Problem 35

Question

Separating variables we obtain $\frac{d y}{(y-1)^{2}+0.01}=d x .$ Then \\[10 \tan ^{-1} 10(y-1)=x+c \text { and } y=1+\frac{1}{10} \tan \frac{x+c}{10}.\\] Setting $x=0$ and $y=1$ we obtain $c=0 .$ The solution is \\[y=1+\frac{1}{10} \tan \frac{x}{10}.\\]

Step-by-Step Solution

Verified

Answer

The solution is $y = 1 + \frac{1}{10} \tan \frac{x}{10}$.

1Step 1: Understanding the Initial Equation

We start with the differential equation $\frac{dy}{(y-1)^2 + 0.01} = dx$. This equation suggests we may be able to separate the variables to solve for $y$ in terms of $x$.

2Step 2: Separating Variables

The given differential form $\frac{dy}{(y-1)^2 + 0.01} = dx$ is already separated. Therefore, integrate the left-hand side with respect to $y$ and the right-hand side with respect to $x$.

3Step 3: Integrating Both Sides

Integrating both sides: For the left side, the integral $\int \frac{dy}{(y-1)^2 + 0.01}$ can be solved as $10 \tan^{-1}(10(y-1))$. For the right side, integrate $dx$ to get $x$. Thus, we obtain the equation $10 \tan^{-1}(10(y-1)) = x + c$.

4Step 4: Solving for y

Solve for $y$ by isolating it: Rearrange $10 \tan^{-1}(10(y-1)) = x + c$ to express $y$ in terms of $x$. We use the inverse tangent function: $y = 1 + \frac{1}{10} \tan\frac{x+c}{10}$.

5Step 5: Applying Initial Conditions

Substitute the initial condition $x = 0$ and $y = 1$ into the equation $y = 1 + \frac{1}{10} \tan\frac{x+c}{10}$. This gives us $1 = 1 + \frac{1}{10} \tan\frac{c}{10}$. Solving this gives $\tan\frac{c}{10} = 0$, resulting in $c = 0$.

6Step 6: Final Solution

Substitute $c = 0$ back into the equation for $y$: This gives the solution $y = 1 + \frac{1}{10} \tan\frac{x}{10}$.

Key Concepts

Separation of VariablesIntegrationInitial ConditionsInverse Trigonometric Functions

Separation of Variables

Separation of variables is a common technique used in solving differential equations. In this method, we manipulate the equation into a form where all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the opposite side. This arrangement allows us to integrate each side separately.

Start with a differential equation like $\frac{dy}{(y-1)^2 + 0.01} = dx$.
Notice how the terms involving $y$ (which includes the differential $dy$ and the expression $(y-1)^2 + 0.01$) are on one side, and the term $dx$ is on the other.
This setup allows us to integrate both sides, a crucial first step in finding our solution.

Once separated, integration becomes possible, and we can proceed to the next stages of solving the equation.

Integration

After separating the variables, the next step is to perform integration. Integration is the process of finding the integral of a function, which is essentially the reverse of differentiation.
For our differential equation:

The left side requires integration with respect to $y$, specifically $\int \frac{dy}{(y-1)^2 + 0.01}$.
This integral can be expressed using an inverse trigonometric function as $10 \tan^{-1}(10(y-1))$.
The right side simply integrates $dx$, resulting in $x$, plus a constant $c$.

Thus, after integrating both sides, we derive The equation $10 \tan^{-1}(10(y-1)) = x + c$. This equation will assist us in determining the specific solution to our differential equation.

Initial Conditions

Initial conditions are additional information about a particular solution of a differential equation, often provided in the form of values for the functions at specific points.
This problem provides the initial conditions $x = 0$ and $y = 1$.

We use these to determine the constant $c$ in our integrated equation $10 \tan^{-1}(10(y-1)) = x + c$.
Substituting $x = 0$ and $y = 1$ yields the equation $1 = 1 + \frac{1}{10} \tan\frac{c}{10}$.
This simplifies to $\tan\frac{c}{10} = 0$, leading to $c = 0$.

Including the initial conditions helps to refine the solution, ensuring it fits the specific scenario described. The initial condition constraints unique adaptation of the constants in the solution.

Inverse Trigonometric Functions

Inverse trigonometric functions are used in this solution to express the integral of the separated differential equation, and they allow us to translate between angular measures and ratios.

The appearance of $\tan^{-1}$ arises because the form $\int \frac{1}{a^2 + x^2}dx$ can be solved by recognizing it as the derivative form of an inverse tangent function.
In this case, the equation $10 \tan^{-1}(10(y-1)) = x + c$ leverages the property that differentiating $\tan^{-1}(x)$ results in $\frac{1}{1 + x^2}$.
Solving the equation for $y$, we use the inverse function $\tan$ to isolate $y$: $y = 1 + \frac{1}{10} \tan\frac{x+c}{10}$.

Using inverse trigonometric functions is integral to interpreting more complex integrals where variable square sums are involved, providing a straightforward means of embedding trigonometric relations within solutions.

Problem 35

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