3.8 Q-1

Question Statement

Find that each of the following equations written against the differential equation is its solution:

1. $x \frac{d y}{d x} = 1 + y$, $y = C x - 1$
2. $x^2(2 y + 1) \frac{d y}{d x} - 1 = 0$, $y^2 + y = C - \frac{1}{x}$
3. $y \frac{d y}{d x} - e^{2 x} = 1$, $y^2 = e^{2 x} + 2 x + 1$
4. $\frac{1}{x} \frac{d y}{d x} - 2 y = 0$, $y = C e^{x^2}$
5. $\frac{d y}{d x} = \frac{y^2 + 1}{e^x}$, $y = \tan(e^x + C)$

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Background and Explanation

To solve these problems, we’ll apply basic concepts of differentiation and verify if the proposed solutions satisfy the differential equations. The method involves:

• Differentiating the proposed solutions with respect to $x$.
• Substituting the derivatives back into the original differential equations.
• Checking if the equation holds true after substitution.

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Solution

1. $x \frac{d y}{d x} = 1 + y$, $y = C x - 1$

We start by differentiating $y = C x - 1$:

$$\frac{d y}{d x} = C$$

Now substitute into the differential equation:

$$x \frac{d y}{d x} = x \cdot C = C x$$

And since $y = C x - 1$, we have:

$$1 + y = 1 + (C x - 1) = C x$$

Thus, $x \frac{d y}{d x} = 1 + y$ holds, confirming that the proposed solution is correct.

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2. $x^2(2 y + 1) \frac{d y}{d x} - 1 = 0$, $y^2 + y = C - \frac{1}{x}$

We start with the proposed solution $y^2 + y = C - \frac{1}{x}$, and differentiate both sides with respect to $x$:

$$2 y \frac{d y}{d x} + \frac{d y}{d x} = 0 + \frac{1}{x^2}$$

Factor the left-hand side:

$$(2 y + 1) \frac{d y}{d x} = \frac{1}{x^2}$$

Now, multiply both sides by $x^2$:

$$x^2 (2 y + 1) \frac{d y}{d x} = 1$$

Thus, we get:

$$x^2 (2 y + 1) \frac{d y}{d x} - 1 = 0$$

This confirms that the proposed solution is correct.

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3. $y \frac{d y}{d x} - e^{2 x} = 1$, $y^2 = e^{2 x} + 2 x + 1$

We start with the proposed solution $y^2 = e^{2 x} + 2 x + 1$, and differentiate both sides with respect to $x$:

$$2 y \frac{d y}{d x} = 2 e^{2 x} + 2$$

Now, solve for $y \frac{d y}{d x}$:

$$y \frac{d y}{d x} = e^{2 x} + 1$$

Substitute this into the original equation:

$$y \frac{d y}{d x} - e^{2 x} = e^{2 x} + 1 - e^{2 x} = 1$$

Thus, the equation holds, confirming the proposed solution.

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4. $\frac{1}{x} \frac{d y}{d x} - 2 y = 0$, $y = C e^{x^2}$

We start with $y = C e^{x^2}$ and differentiate with respect to $x$:

$$\frac{d y}{d x} = 2 x C e^{x^2}$$

Now substitute into the original equation:

$$\frac{1}{x} \cdot 2 x C e^{x^2} - 2 C e^{x^2} = 2 C e^{x^2} - 2 C e^{x^2} = 0$$

Thus, the equation holds, confirming the proposed solution.

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5. $\frac{d y}{d x} = \frac{y^2 + 1}{e^x}$, $y = \tan(e^x + C)$

We start with $y = \tan(e^x + C)$ and differentiate with respect to $x$:

$$\frac{d y}{d x} = \sec^2(e^x + C) \cdot e^x$$

Since $\sec^2(z) = 1 + \tan^2(z)$, we have:

$$\frac{d y}{d x} = (1 + y^2) e^x$$

Now, substitute into the original equation:

$$\frac{d y}{d x} = \frac{y^2 + 1}{e^x}$$

Thus, the equation holds, confirming the proposed solution.

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Key Formulas or Methods Used

• Differentiation: Used to find $\frac{d y}{d x}$ from the proposed solutions.
• Substitution: Substituted the derivatives into the original differential equations to check if they hold.
• Basic trigonometric identities: Used for the tangent and secant functions.

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Summary of Steps

1. Differentiate the proposed solution with respect to $x$.
2. Substitute the derivative back into the original differential equation.
3. Simplify the equation and verify if it holds true.
4. Confirm the correctness of the solution by checking if both sides of the equation are equal.