Problem 3 - Advanced Engineering Mathematics

Question

In Problems, solve the given differential equation by using the substitution \(u=y^{\prime}\). $$ y^{\prime \prime}+\left(y^{\prime}\right)^{2}+1=0 $$

Step-by-Step Solution

Answer

The solution is \( y = -\ln|\cos(C-x)| + K \).

Step 1: Substitute u for y'

Use the substitution \( u = y' \), which implies \( y'' = u' \). The given differential equation \( y'' + (y')^2 + 1 = 0 \) becomes: \( u' + u^2 + 1 = 0 \).

Step 2: Solve the new differential equation for u

Rewrite the differential equation in terms of \( u \): \( u' = -u^2 - 1 \). This can be rewritten as \( \frac{du}{dx} = -u^2 - 1 \).

Step 3: Separate the variables for u

Separate the variables: \( \frac{du}{-u^2 - 1} = dx \). Integrate both sides to find an expression involving \( u \) and \( x \).

Step 4: Integrate both sides

The left side becomes an integral: \( \int \frac{du}{-u^2 - 1} = \int dx \). The integral of \( \frac{1}{-u^2-1} \) is \( -\tan^{-1}(u) \). On the right side, the integral is \( x + C \), where \( C \) is the integration constant. Thus, \( -\tan^{-1}(u) = x + C \).

Step 5: Solve for u

Solve \( -\tan^{-1}(u) = x + C \) for \( u \): \( u = \tan(-(x + C)) \) or \( u = \tan(C - x) \).

Step 6: Resubstitute u back to y'

Remember \( u = y' \), so \( y' = \tan(C - x) \). This will allow us to find the function \( y \) by integrating with respect to \( x \).

Step 7: Integrate to find y

Integrate \( y' = \tan(C - x) \) to find \( y \): \( y = \int \tan(C - x) \, dx \). This yields \( y = -\ln|\cos(C-x)| + K \), where \( K \) is another constant of integration.

Key Concepts

Substitution Method

When dealing with complex differential equations, substitution is a handy tool. Imagine you have a messy equation involving multiple terms. The substitution method simplifies the equation by replacing a variable or an expression with something more manageable. In our example, the substitution involved is setting \( u = y' \). This means we're renaming the derivative of \( y \) as \( u \).

• Why use substitution? It turns our original problem into something potentially easier to handle.
• We also change notation: if \( y' = u \), then \( y''=u' \).

With these changes, the equation \( y'' + (y')^2 + 1 = 0 \) transforms into \( u' + u^2 + 1 = 0 \). See how much simpler it looks? This technique is a foundational tool in solving differential equations more efficiently.

Second-Order Differential Equations

Second-order differential equations involve the second derivative of a function. The form can look tricky because you're dealing with higher levels of differentiation. For instance, our original equation is \( y'' + (y')^2 + 1 = 0 \). Here, \( y'' \) is the second derivative of \( y \).

• The presence of \( y'' \) typically makes the equation more challenging to solve.
• These types of equations often require specialized methods like the substitution method we've discussed.

Understanding how to tackle these higher-order derivatives is essential for mastering differential equations. Once simplified, such equations can become manageable with proper techniques.

Variable Separation

Variable separation is like an organizational tool in math. It's a method where you arrange an equation so that all terms containing one variable are on one side and terms with another are on the opposite side. In our example, after substitution, the equation \( u' = -u^2 - 1 \) is separated into: \[ \frac{du}{-u^2 - 1} = dx \]

• This step is crucial because it allows each side of the equation to be integrated separately.
• It's often necessary to ensure the equation is manipulable into this separable form.

Once separated, you're ready to integrate each side, tackling one variable at a time! This approach streamlines solving differential equations, turning it into a step-by-step process.

Integration Techniques

Integration is about finding the antiderivative of a function. In the context of differential equations, it's the phase where you solve for the function in terms of its independent variable, often \( x \). For the exercise's equation, we've reached a point where integration of both sides is needed.

• The left-hand side: \( \int \frac{du}{-u^2 - 1} \).
• The right-hand side simply features the integral \( \int dx \).

The solution involves familiar techniques. For \( \frac{du}{-u^2 - 1} \), recognizing the function is similar to the derivative of \( \tan^{-1}(u) \) helps in integration. The integration results in: \[ -\tan^{-1}(u) = x + C \] This step is crucial as it connects \( u \) back to \( x \) and helps in resubstituting to find \( y \). Remember, integration will also introduce constants of integration, reflecting the indefinite nature of these solutions.