Find The General Solution Of The Differential Equation (x^2 + 1)tan Y Dy/dx = X. (a) Y = C/squareroot

Find The General Solution Of The Differential Equation (x^2 + 1)tan Y Dy/dx = X. (a) Y = C/squareroot

Understanding and solving differential equations is a fundamental aspect of advanced mathematics, with applications spanning physics, engineering, and other scientific disciplines. In this article, we focus on a specific type of differential equation:

\[
(x^2 + 1) \tan Y \frac{dy}{dx} = x
\]

Our goal is to find its general solution, especially under the condition \( Y = \frac{C}{\sqrt{x}} \). Through systematic steps, substitution methods, and integration techniques, we will explore how to approach and solve this equation comprehensively.

---

Introduction to the Differential Equation

The differential equation in question is:

\[
(x^2 + 1) \tan Y \frac{dy}{dx} = x
\]

This is a first-order differential equation involving a nonlinear term \(\tan Y\). The equation can be viewed as a relationship between the dependent variable \(Y\) and the independent variable \(x\), where the derivative \(dy/dx\) is multiplied by a function involving \(x\) and \(Y\).

Key features of this differential equation include:

• Nonlinear dependence on \(Y\) via \(\tan Y\).


• The coefficient \((x^2 + 1)\), which is always positive.


• The right-hand side is a simple function \(x\).

Understanding the Structure of the Equation

Before attempting to solve, it's essential to analyze the structure:

• The equation resembles a separable form if we can express it in terms of \(Y\) and \(x\) separately.


• The presence of \(\tan Y\) suggests that substitution involving the tangent function might simplify the problem.


• The form hints at a substitution \(Z = \tan Y\), which could linearize or simplify the equation.

Methodology for Solving the Differential Equation

The solution approach involves several key steps:

1. Substitution to Simplify the Equation


2. Rearrangement to a Separable Form


3. Integration of Both Sides


4. Applying Initial Conditions or Given Conditions


5. Expressing the General Solution

---

Step 1: Substitution to Simplify \(\tan Y\)

Since \(\tan Y\) appears prominently, define a new variable:

\[
Z = \tan Y
\]

Then, using the derivative chain rule:

\[
\frac{dy}{dx} = \frac{1}{\sec^2 Y} \frac{dZ}{dx}
\]

But this adds complexity, so a more straightforward approach is to directly rewrite the original differential equation in terms of \(Z\):

\[
(x^2 + 1) Z \frac{dy}{dx} = x
\]

Recall that:

\[
Y = \arctan Z
\]

Differentiating both sides with respect to \(x\):

\[
\frac{dY}{dx} = \frac{1}{1 + Z^2} \frac{dZ}{dx}
\]

Express \(dy/dx\) in terms of \(Z\):

\[
\frac{dy}{dx} = \frac{1}{1 + Z^2} \frac{dZ}{dx}
\]

Substitute into the original equation:

\[
(x^2 + 1) Z \cdot \frac{1}{1 + Z^2} \frac{dZ}{dx} = x
\]

---

Step 2: Formulating the Equation in Terms of \(Z\)

Rearranging:

\[
\frac{(x^2 + 1) Z}{1 + Z^2} \frac{dZ}{dx} = x
\]

Expressed as:

\[
\frac{dZ}{dx} = \frac{x (1 + Z^2)}{(x^2 + 1) Z}
\]

This form is more manageable because it separates variables \(Z\) and \(x\):

\[
\frac{dZ}{(1 + Z^2)/Z} = \frac{x}{x^2 + 1} dx
\]

Note that:

\[
\frac{1 + Z^2}{Z} = Z + \frac{1}{Z}
\]

Therefore, the differential equation becomes:

\[
\frac{dZ}{Z + \frac{1}{Z}} = \frac{x}{x^2 + 1} dx
\]

Simplify the left side:

\[
\frac{dZ}{Z + \frac{1}{Z}} = \frac{dZ}{\frac{Z^2 + 1}{Z}} = \frac{Z}{Z^2 + 1} dZ
\]

Thus, the equation reduces to:

\[
\frac{Z}{Z^2 + 1} dZ = \frac{x}{x^2 + 1} dx
\]

---

Step 3: Integration of Both Sides

Now, integrate both sides:

\[
\int \frac{Z}{Z^2 + 1} dZ = \int \frac{x}{x^2 + 1} dx
\]

Left Integral:

Let’s evaluate:

\[
I_1 = \int \frac{Z}{Z^2 + 1} dZ
\]

Use substitution:

• Set \(u = Z^2 + 1\),


• Then \(du = 2Z dZ\),


• So, \(Z dZ = \frac{1}{2} du\).

\[
I1 = \int \frac{Z}{Z^2 + 1} dZ = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| + C1
\]

Back to \(Z\):

\[
I1 = \frac{1}{2} \ln |Z^2 + 1| + C1
\]

Right Integral:

Similarly,

\[
I_2 = \int \frac{x}{x^2 + 1} dx
\]

Let \(v = x^2 + 1\), then \(dv = 2x dx\), so:

\[
x dx = \frac{1}{2} dv
\]

Thus,

\[
I2 = \frac{1}{2} \int \frac{1}{v} dv = \frac{1}{2} \ln |v| + C2 = \frac{1}{2} \ln |x^2 + 1| + C_2
\]

---

Step 4: Combining the Results

The integrated form is:

\[
\frac{1}{2} \ln |Z^2 + 1| = \frac{1}{2} \ln |x^2 + 1| + C
\]

or

\[
\ln |Z^2 + 1| = \ln |x^2 + 1| + 2C
\]

Let \(K = e^{2C}\), a constant, then:

\[
|Z^2 + 1| = K |x^2 + 1|
\]

Since \(Z = \tan Y\), substitute back:

\[
(\tan Y)^2 + 1 = K (x^2 + 1)
\]

Recall the Pythagorean identity:

\[
(\tan Y)^2 + 1 = \sec^2 Y
\]

Therefore:

\[
\sec^2 Y = K (x^2 + 1)
\]

---

Step 5: Expressing the General Solution

From the above, we have:

\[
\sec^2 Y = K (x^2 + 1)
\]

Taking square roots:

\[
\sec Y = \pm \sqrt{K} \sqrt{x^2 + 1}
\]

Expressing \(Y\):

\[
Y = \pm \arccos \left( \frac{1}{\sqrt{K} \sqrt{x^2 + 1}} \right) + n\pi
\]

where \(n\) is an integer, accounting for the periodicity of \(\arccos\).

Alternatively, the solution can be written in terms of tangent:

\[
\tan Y = \pm \sqrt{K (x^2 + 1) - 1}
\]

Since the problem specifies a particular solution form:

Given Condition:

\[
Y = \frac{C}{\sqrt{x}}
\]

we can relate the constant \(K\) to \(C\) to satisfy this initial condition or particular solution form.

---

Special Case: Y = C / \(\sqrt{x}\)

Suppose \(Y = \frac{C}{\sqrt{x}}\):

• The general solution involves \(\sec^2 Y\), which becomes:

-