Problem 13 - Thomas Calculus

Question

Solve the differential equation in Exercises \(9-18\). $$ \frac{d y}{d x}=\sqrt{y} \cos ^{2} \sqrt{y} $$

Step-by-Step Solution

Answer

The solution is \( y = \left(\tan^{-1}\left(\frac{x + C}{2}\right)\right)^2 \).

Step 1: Separate Variables

Rewrite the differential equation \( \frac{dy}{dx} = \sqrt{y} \cos^2(\sqrt{y}) \) by separating variables. Move terms involving \(y\) to the left side and \(x\) to the right side: \( \frac{dy}{\sqrt{y} \cos^2(\sqrt{y})} = dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation. Solve \( \int \frac{dy}{\sqrt{y} \cos^2(\sqrt{y})} \) on the left side and \( \int dx \) on the right side. Make a substitution to simplify the left side: let \( u = \sqrt{y} \) which implies \( du = \frac{1}{2\sqrt{y}} dy \).

Step 3: Substitution and Simplification

Substitute \( du = \frac{1}{2\sqrt{y}} dy \) into the integral. The left side becomes \( 2 \int \frac{1}{\cos^2(u)} du = 2 \int \sec^2(u) du \).

Step 4: Solve Integrals

Solve the integral \( 2 \int \sec^2(u) du \) which is \( 2 \tan(u) + C_1 \) where \(C_1\) is a constant of integration. The right side integration gives \( x + C_2 \) where \(C_2\) is another constant of integration.

Step 5: Combine Solutions

Combine the solutions from Step 4: \( 2 \tan(\sqrt{y}) = x + C \), where \(C = C_2 - C_1\).

Step 6: Solve for y

To find \(y\), solve the equation for \(\sqrt{y}\) and then square it to find \(y\). Start by isolating \(\tan(\sqrt{y})\): \( \tan(\sqrt{y}) = \frac{x + C}{2} \), then \( \sqrt{y} = \tan^{-1}\left(\frac{x + C}{2}\right) \), and finally \( y = \left(\tan^{-1}\left(\frac{x + C}{2}\right)\right)^2 \).

Key Concepts

Separation of Variables

Separation of variables is a powerful method for solving ordinary differential equations. This technique involves rearranging the equation so that each variable (typically, one is dependent and the other independent) is isolated on different sides of the equation. By separating variables, we make it easier to integrate both sides to find the solution.

In the provided differential equation, \( \frac{dy}{dx} = \sqrt{y} \cos^2(\sqrt{y}) \), the strategy was to separate terms involving \(y\) and \(x\). The goal is to express the equation in a format like \( g(y) \, dy = h(x) \, dx \), enabling integration.

• Move \(\sqrt{y} \cos^2(\sqrt{y})\) to the left side with \(dy\).
• Keep \(dx\) on the right side.

After this arrangement, the integrals \( \int \frac{dy}{\sqrt{y} \cos^2(\sqrt{y})} \) and \( \int dx \) can be solved separately, typically requiring further simplification or substitution.

Integration Techniques

In solving differential equations, integration techniques are crucial. After separating variables, the key task is to integrate both sides of the equation. For our example, we encountered the integral \( \int \frac{dy}{\sqrt{y} \cos^2(\sqrt{y})} \).

Here, the use of substitution becomes quite handy. By letting \( u = \sqrt{y} \), we express \( dy \) in terms of \( du \): \( dy = 2u \, du \). This converts the original integral into a more standard form:

• \( 2 \int \sec^2(u) \, du \) is easily recognized and solvable using integration rules.
• The integral of \( \sec^2(u) \) is \( \tan(u) + C \).
• This integration reflects the trigonometric function commonly apparent in calculus.

Thus, integration techniques, combined with substitutions, transform complex expressions into simpler forms.

Trigonometric Substitution

Trigonometric substitution is a method used to solve integrals, particularly those involving square roots and trigonometric identities. This technique can simplify otherwise challenging integrals.

In the given differential equation, trigonometric substitution plays a crucial role in simplifying \( \int \frac{dy}{\sqrt{y} \cos^2(\sqrt{y})} \) into \( 2 \int \sec^2(u) \, du \).

• By replacing \( \sqrt{y} \) with \( u \) as suggested, our integral now entails trigonometric functions of \( u \).
• The \( \sec^2(u) \) term is manageable, having a direct antiderivative: \( \tan(u) \).

The strategic use of trigonometric substitution simplifies the integration process, ultimately leading us back from \( u \) to the original variable, \( y \), through inverse trigonometric functions like \( \tan^{-1}(x) \), making it easier to solve and interpret the solution of the differential equation.