Let slope of the tangent line to a curve at any point P(x, y) be given by (xy²+y)/x. If the curve in

Q: Wait, why can't I just separate the variables $x$ and $y$ at the very beginning?

That is a great instinct, but look closely at the term $\frac{xy^2+y}{x}$. If you try to separate them, you will find that the $y^2$ and $y$ terms are 'coupled' with $x$ in a way that prevents you from isolating all $y$ terms on one side and all $x$ terms on the other. This is exactly why we need the Bernoulli transformation—it is designed specifically to decouple these variables.

Q: How do I know when to use the Bernoulli substitution $v = y^{1-n}$?

Whenever you see a differential equation where the right-hand side is a polynomial in $y$ (like $y^2 + y/x$), it is a huge red flag for Bernoulli's form. If you see a $\frac{dy}{dx}$ term, a linear $y$ term, and a $y^n$ term (where $n \neq 0, 1$), that is your cue. The substitution $v = y^{1-n}$ is the magic key that turns that non-linear $y^n$ term into a constant, linearizing the whole equation.

Q: Why did we get a negative sign when differentiating $v = y^{-1}$?

That is the chain rule in action! When we differentiate $v = y^{-1}$ with respect to $x$, we treat $y$ as a function of $x$. So, $\frac{dv}{dx} = \frac{d}{dy}(y^{-1}) \cdot \frac{dy}{dx}$, which gives $-y^{-2} \cdot \frac{dy}{dx}$. That negative sign is crucial—it is what allows us to cancel out the terms and arrive at the standard linear form.

Question

Let slope of the tangent line to a curve at any point P(x, y) be given by $\frac{xy^2+y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$, then the value of y, for which the point (3, y) lies on the curve, is :

Options:

(A) $\frac{18}{11}$

(B) $\frac{18}{19}$

(C) $-\frac{4}{3}$

(D) $\frac{18}{35}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a curve. You don't know its equation, but you know its secret: the slope of the tangent line at any point $P(x, y)$ is defined by the expression:

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

This is not just a math problem; it is a story of transformation. We are going to take this complex, non-linear relationship and, through a series of logical steps, reveal the true identity of the curve.

### Phase 1: The Simplification

Before we panic, let's look at the expression again. The right-hand side is begging to be broken apart. If we divide each term in the numerator by the denominator $x$, the equation becomes much friendlier:

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

Now, let's bring the $y$ term to the left side to see what we are really dealing with:

$$\frac{dy}{dx} - \frac{1}{x}y = y^2$$

This is the classic **Bernoulli differential equation**. It is non-linear because of that $y^2$ term, which prevents us from using standard linear techniques.

### Phase 2: The Bernoulli Transformation

To linearize this, we need a clever substitution. We divide the entire equation by $y^2$ to isolate the non-linear part:

$$\frac{1}{y^2}\frac{dy}{dx} - \frac{1}{xy} = 1$$

Now, let's define a new variable $v = y^{-1}$. When we differentiate this with respect to $x$ using the chain rule, we get $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$.

Substituting this back in, we get:

$$-\frac{dv}{dx} - \frac{1}{x}v = 1$$

Multiplying by $-1$ gives us the standard linear form we love:

$$\frac{dv}{dx} + \frac{1}{x}v = -1$$

### Phase 3: The Power of the Integrating Factor

We are now in familiar territory. This is a first-order linear differential equation. To solve it, we need an Integrating Factor (I.F.), defined by $e^{\int P(x) dx}$, where $P(x) = \frac{1}{x}$.

$$I.F. = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$$

Multiplying our linear equation by this integrating factor $x$ makes the left side a perfect derivative of the product $v \cdot x$:

$$\frac{d}{dx}(vx) = -x$$

Integrating both sides with respect to $x$ yields:

$$vx = \int -x dx + C = -\frac{x^2}{2} + C$$

### Phase 4: Finding the Identity

We are almost there. Substituting $v = \frac{1}{y}$ back into our equation, we find the general family of curves:

$$\frac{x}{y} = -\frac{x^2}{2} + C$$

But we need the specific curve that intersects the line $x + 2y = 4$ at $x = -2$. First, let's find the point of intersection by plugging $x = -2$ into the line equation: $-2 + 2y = 4$, which means $2y = 6$, so $y = 3$.

Our curve passes through the point $(-2, 3)$. Now, we solve for $C$:

$$\frac{-2}{3} = -\frac{(-2)^2}{2} + C \Rightarrow -\frac{2}{3} = -2 + C \Rightarrow C = 2 - \frac{2}{3} = \frac{4}{3}$$

### Phase 5: The Final Calculation

With $C = \frac{4}{3}$, the equation of our curve is $\frac{x}{y} = -\frac{x^2}{2} + \frac{4}{3}$. The problem asks for the value of $y$ when $x = 3$. Let's plug it in:

$$\frac{3}{y} = -\frac{3^2}{2} + \frac{4}{3} = -\frac{9}{2} + \frac{4}{3}$$

Finding a common denominator of 6:

$$\frac{3}{y} = \frac{-27 + 8}{6} = -\frac{19}{6}$$

Inverting both sides, we get the final result:

**$y = -\frac{18}{19}$**

Animated Solution for Mathematics - Differential Equations: Let slope of the tangent line to a curve at any point P(x, y) be given by $\frac{x y ^{2} + y}{x}$ . If the curve intersects the line $x + 2 y = 4$ at $x = - 2$ , then the value of y, for which the point (3, y) lies on the curve, is :

Select Answer:

Visualized Solution

Identifying the Slope Equation

Simplifying the Expression

Recognizing Bernoulli's Form

Linearizing the Equation

Differentiating the Substitution

The Linear Differential Equation

Calculating the Integrating Factor

Finding the General Solution

Reverting the Substitution

The Intersection Point

Solving for Constant C

The Final Curve Equation

Finding y for x = 3

Final Calculation

Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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