If a curve passes through the point (1, -2) and has slope of the tangent at any point (x, y) on it a

Q: Wait, why can't I just separate the variables and integrate directly?

That is a very common temptation! Separation of variables only works if you can write the equation as $f(x)dx = g(y)dy$. Here, the term $\frac{2y}{x}$ couples $x$ and $y$ in a way that prevents simple separation. When you see a term like $\frac{dy}{dx} + P(x)y = Q(x)$, separation is not the right tool—the Integrating Factor is.

Q: How do I know when to use the Integrating Factor method?

Look for the 'Linearity' signature. If you see a $y$ and a $\frac{dy}{dx}$ appearing linearly (meaning they aren't squared, inside a sine function, or multiplied together), and the rest of the equation depends only on $x$, you are almost certainly looking at a Linear Differential Equation. It is a pattern recognition skill that you will master with practice.

Q: Why did we reject the negative root or other options?

We didn't reject them arbitrarily! We found the unique equation of the curve that passes through $(1, -2)$. Once we have the specific equation $4yx^2 = x^4 - 9$, any point that lies on the curve must satisfy this equality. We tested the options by plugging them into the equation, and $(\sqrt{3}, 0)$ was the only one that balanced the equation perfectly.

Question

If a curve passes through the point $(1, -2)$ and has slope of the tangent at any point $(x, y)$ on it as $\frac{x^2 - 2y}{x}$, then the curve also passes through the point :

Options:

(A) $(\sqrt{2}, 1)$

(B) $(\sqrt{3}, 0)$

(C) $(-1, 2)$

(D) $(3, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a path defined by a curve. You do not know the equation of this path, but you know something fundamental about how it behaves: at any point $(x, y)$, the slope of the tangent is given by:

$$ \frac{dy}{dx} = \frac{x^2 - 2y}{x} $$

We are given a starting point, $(1, -2)$, which acts as our anchor in the coordinate plane. Our goal is to find the identity of this curve.

### The Transformation: Recognizing the Pattern

When we first look at the equation, it feels slightly chaotic. However, we can break the fraction apart to reveal a hidden structure:

$$ \frac{dy}{dx} = \frac{x^2}{x} - \frac{2y}{x} \Rightarrow \frac{dy}{dx} = x - \frac{2y}{x} $$

Now, let us bring the $y$ term to the left side to obtain:

$$ \frac{dy}{dx} + \frac{2}{x}y = x $$

Suddenly, the fog clears. This is the classic **Linear Differential Equation** of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2}{x}$ and $Q(x) = x$.

### The Magic of the Integrating Factor

To solve this, we need a special tool: the Integrating Factor (I.F.). Think of the I.F. as a mathematical catalyst that forces the left side of our equation to become the derivative of a product.

The formula is defined as:

$$ I.F. = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} $$

Since the integral of $\frac{2}{x}$ is $2 \ln(x)$, we use the properties of logarithms to simplify:

$$ I.F. = e^{2 \ln(x)} = e^{\ln(x^2)} = x^2 $$

This result, $x^2$, is the key that unlocks the door to the solution.

### The Integration Journey

Now, we multiply our entire differential equation by this Integrating Factor:

$$ x^2 \left( \frac{dy}{dx} + \frac{2}{x}y \right) = x^2(x) $$

This simplifies to:

$$ x^2 \frac{dy}{dx} + 2xy = x^3 $$

Look closely at the left side; it is exactly the derivative of the product $y \cdot x^2$. Thus, we can write:

$$ \frac{d}{dx}(y x^2) = x^3 $$

Integrating both sides with respect to $x$ yields:

$$ \int \frac{d}{dx}(y x^2) dx = \int x^3 dx \Rightarrow y x^2 = \frac{x^4}{4} + C $$

### Finding the Anchor

We have the general family of curves, but we need the specific one that passes through $(1, -2)$. Substituting $x = 1$ and $y = -2$ into our equation:

$$ (-2)(1)^2 = \frac{(1)^4}{4} + C $$

This leads to $-2 = \frac{1}{4} + C$, which means $C = -2 - \frac{1}{4} = -\frac{9}{4}$. Our specific curve equation is:

$$ y x^2 = \frac{x^4}{4} - \frac{9}{4} $$

Multiplying by 4 to clear the fractions, we arrive at the final equation:

$$ **4 y x^2 = x^4 - 9** $$

### The Final Verification

Finally, we test the point $(\sqrt{3}, 0)$. Substituting $x = \sqrt{3}$ and $y = 0$ into our derived equation:

$$ 4(0)(\sqrt{3})^2 = (\sqrt{3})^4 - 9 $$

This simplifies to $0 = 9 - 9$, which is $0 = 0$. The point satisfies the equation perfectly, confirming our result.

Animated Solution for Mathematics - Differential Equations: If a curve passes through the point $(1, - 2)$ and has slope of the tangent at any point $(x, y)$ on it as $\frac{x ^{2} - 2 y}{x}$ , then the curve also passes through the point :

Select Answer:

Visualized Solution

Understanding the Slope Condition

Rearranging to Linear Form

Calculating the Integrating Factor

Setting up the General Solution

Integrating and Finding the Constant

The Final Equation of the Curve

Testing the Options

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation

The Magic of the Integrating Factor

The Integration Journey

Finding the Anchor

The Final Verification

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