If the solution curve of the differential equation (2x - 10y³) dy + y dx = 0, passes through the poi

Q: Wait, why do we use $dx/dy$ instead of $dy/dx$? Isn't $dy/dx$ the standard?

That is a fantastic question! While $dy/dx$ is indeed the most common form, differential equations are all about flexibility. If you try to force this into $dy/dx$, you will end up with a non-linear mess. By observing that $x$ only appears to the power of 1, we realize that treating $x$ as a function of $y$ makes the equation linear, which is much easier to solve. Always look for the variable that appears linearly!

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

You use an Integrating Factor whenever you can transform your differential equation into the standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$ or $\frac{dx}{dy} + P(y)x = Q(y)$. If you see a term like $y'$ and a term with $y$ (or $x'$ and $x$), that is your signal to use the IF method. It is your most powerful tool for solving first-order linear equations.

Q: Why does the constant $C$ matter so much?

The constant $C$ represents the entire family of curves that satisfy the differential equation. Without it, you only have a general solution. By using the given point $(0, 1)$, we 'pin' the curve down to one specific path. It is the difference between knowing the general behavior of a system and knowing exactly where it is at a specific moment.

Question

If the solution curve of the differential equation $(2x - 10y^3) dy + y dx = 0$, passes through the points $(0, 1)$ and $(2, \beta)$, then $\beta$ is a root of the equation:

Options:

(A) $y^5 - 2y - 2 = 0$

(B) $2y^5 - 2y - 1 = 0$

(C) $2y^5 - y^2 - 2 = 0$

(D) $y^5 - y^2 - 1 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, winding path—a differential equation that seems to lead nowhere. The equation $(2x - 10y^3) dy + y dx = 0$ might look intimidating at first glance, but in the world of JEE Advanced, every equation is a puzzle waiting for the right perspective.

The secret here isn't brute force; it is the art of transformation. We are looking for a curve, a specific path in the $xy$-plane. Before we can find it, we must rewrite the equation in a language we understand.

### Rearranging the Chaos

Most students instinctively reach for $\frac{dy}{dx}$. However, notice that $x$ appears only to the first power. That is a massive, glowing neon sign telling us to switch our perspective.

Let us rearrange the terms to isolate $\frac{dx}{dy}$:

$$y \frac{dx}{dy} = 10y^3 - 2x$$

Now, let us bring the $x$ terms to one side:

$$y \frac{dx}{dy} + 2x = 10y^3$$

Finally, divide by $y$ to reach the standard linear form:

$$\frac{dx}{dy} + \frac{2}{y}x = 10y^2$$

Suddenly, the chaos has order. We are now looking at a classic linear differential equation in the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \frac{2}{y}$ and $Q(y) = 10y^2$.

### The Magic Key: The Integrating Factor

Every linear differential equation has a 'magic key' that unlocks its solution: the Integrating Factor (IF). The formula is defined as:

$$IF = e^{\int P(y) dy}$$

Let us plug in our $P(y)$:

$$IF = e^{\int \frac{2}{y} dy}$$

The integral of $\frac{2}{y}$ is $2 \ln y$, which, by the laws of logarithms, is $\ln y^2$. So, our IF becomes $e^{\ln y^2}$. Because $e$ and $\ln$ are inverse functions, they cancel out perfectly, leaving us with:

$$IF = y^2$$

This is the moment where the math starts to feel elegant. We have found the key that will simplify everything.

### The Path to the Solution

The general solution for our linear equation is given by $x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$. Substituting our values, we get:

$$x \cdot y^2 = \int (10y^2) \cdot y^2 dy + C$$

This simplifies to:

$$x y^2 = \int 10y^4 dy + C$$

Integrating $10y^4$ is a simple application of the power rule, giving us $10 \cdot \frac{y^5}{5} = 2y^5$. Thus, our general solution is:

$$x y^2 = 2y^5 + C$$

### Pinning Down the Curve

We have a family of curves, but the problem states the curve passes through $(0, 1)$. Let us use this to find our constant $C$.

Substituting $x = 0$ and $y = 1$ into our equation:

$$0 \cdot (1)^2 = 2(1)^5 + C \Rightarrow 0 = 2 + C \Rightarrow C = -2$$

Our specific curve is $x y^2 = 2y^5 - 2$. Finally, we are told the curve also passes through $(2, \beta)$.

Substituting $x = 2$ and $y = \beta$, we get:

$$2 \beta^2 = 2 \beta^5 - 2$$

Dividing by $2$, we get $\beta^2 = \beta^5 - 1$, or:

$$\beta^5 - \beta^2 - 1 = 0$$

We have arrived at the final destination. **$\beta$ is a root of the polynomial $y^5 - y^2 - 1 = 0$.** You have successfully navigated the differential landscape!

Animated Solution for Mathematics - Differential Equations: If the solution curve of the differential equation $(2 x - 10 y^{3}) d y + y d x = 0$ , passes through the points $(0, 1)$ and $(2, β)$ , then $β$ is a root of the equation:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging the Terms

Standard Linear Form in $x$

Identify $P (y)$ and $Q (y)$

Calculating the Integrating Factor

Simplifying the I.F.

The General Solution Formula

Solving the Integral

Finding the Constant $C$

The Specific Curve Equation

Substituting the Point $(2, β)$

Deriving the Equation for $β$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Rearranging the Chaos

The Magic Key

The Path to the Solution

Pinning Down the Curve

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