91Ó°ÊÓ

Separating variables is a powerful technique when dealing with differential equations. It involves rearranging an equation so that each variable and its differentials are on opposite sides of the equation. When a differential equation can be expressed in the "dy/dx = g(x)h(y)" form, it suggests that separation is possible. In our exercise, we start with \( (1 + x^4) dy + x(1 + 4y^2) dx = 0 \). By dividing and rearranging, we expressed it as \( \frac{dy}{1 + 4y^2} = -\frac{x}{1 + x^4} dx \). This shows variables exclusively on each side: - The left side depends only on \( y \) and \( dy \). - The right side depends only on \( x \) and \( dx \).
This clear separation allows us to integrate each side independently.

An Initial Value Problem (IVP) is a differential equation accompanied by a condition that allows finding a specific solution. The initial condition enables the determination of any integration constants that appear during the solution process. In our case, the initial condition is given as \( y(1) = 0 \).

• During the solution process, we integrate both sides of the separated equation resulting in arbitrarily constant terms \( C_1 \) and \( C_2 \).
• The initial condition, \( y(1) = 0 \), is applied to our integrated equation to find the constant \( C \).
• Substituting \( x = 1 \) and \( y = 0 \) into our integrated equation, ensures the output satisfies the problem’s prescribed starting point.

This approach narrows the infinite number of possible solutions down to one specific solution that fits the given condition.

Integration is the process of finding the integral of a function, which is the inverse operation of differentiation. Here, integration is employed to find the explicit form of the solution after separating variables.

• For the left side, the integral \( \int \frac{dy}{1 + 4y^2} \) correlates with the inverse tangent function, yielding \( \frac{1}{2} \tan^{-1}(2y) + C_1 \).
• On the right side, \( \int -\frac{x}{1 + x^4} dx \) is tackled using a substitution method with \( u = x^2 \), which simplifies the integration to \( -\frac{1}{2} \tan^{-1}(x^2) + C_2 \).

Integration transforms an abstract differential equation into a more manageable expression, enabling the derivation of an explicit function that relates \( y \) with \( x \).

Trigonometric substitution is a technique often used in integration to simplify an expression into a basic trigonometric identity. In solving our differential equation, after separating variables and setting up the integration expression:

• The integral \( \int \frac{dy}{1 + 4y^2} = \frac{1}{2} \tan^{-1}(2y) + C_1 \) directly involves the inverse trigonometric function \( \tan^{-1} \).
• This is used because substituting \( 2y \) into the arctangent simplifies the process, matching a known integral formula \( \int \frac{du}{a^2+u^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C \).

Trigonometric identities not only simplify the integration but also result in a form that makes addressing the initial conditions more manageable. In our problem, using trigonometric substitution made the process not only possible but relatively straightforward, converting the differential equation to an explicit solution format.