JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: The general solution of the differential equation $(y^{2} - x^{3}) d x - x y d y = 0$ ( $x \neq = 0$ ) is :

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look closely at the structure of the differential equation

(y^{2} - x^{3}) d x - x y d y = 0

. It is not immediately separable, nor is it a standard linear equation. Notice the presence of

y^{2}

and

x^{3}

. This structure is a classic indicator of a Bernoulli differential equation. Your goal is to transform this into a linear form by choosing an appropriate substitution.

Hint 2

Hint 2 (The Arsenal): To linearize the equation, you need to eliminate the non-linear

y

terms. Consider the substitution

k = y^{2}

. When you differentiate this with respect to

x

, you will obtain an expression involving

y \frac{d y}{d x}

. This will allow you to replace the

y \frac{d y}{d x}

term in your rearranged equation, effectively removing the non-linearity.

Hint 3

Hint 3 (The Execution): Start by rearranging the original equation to isolate

\frac{d y}{d x}

, which gives

\frac{d y}{d x} = \frac{y ^{2} - x ^{3}}{x y}

. Then, differentiate

k = y^{2}

to get

\frac{d k}{d x} = 2 y \frac{d y}{d x}

. Substitute these into the equation to arrive at the linear form

\frac{d k}{d x} - \frac{2}{x} k = - 2 x^{2}

. Now, solve this using the integrating factor method.

Visualized Solution

Analyze the Differential Equation

Given equation: $(y^{2} - x^{3}) d x - x y d y = 0$
Constraint: $x \neq = 0$

Rearranging for $\frac{d y}{d x}$

Rearranging the terms: $x y d y = (y^{2} - x^{3}) d x$
Expressing as a derivative: $\frac{d y}{d x} = \frac{y ^{2} - x ^{3}}{x y}$

Identifying the Substitution

Let $y^{2} = k$
This substitution aims to linearize the equation in terms of $k$ .

Differentiating the Substitution

Differentiating $y^{2} = k$ with respect to $x$ :
Using Chain Rule: $2 y \frac{d y}{d x} = \frac{d k}{d x}$
Therefore, $y \frac{d y}{d x} = \frac{1}{2} \frac{d k}{d x}$

Substituting back into the Equation

Original rearranged form: $x y \frac{d y}{d x} = y^{2} - x^{3}$
Substituting $y^{2} = k$ and $y \frac{d y}{d x} = \frac{1}{2} \frac{d k}{d x}$ :
Result: $x (\frac{1}{2} \frac{d k}{d x}) = k - x^{3}$

Transforming to Linear Form

Multiplying by $\frac{2}{x}$ : $\frac{d k}{d x} = \frac{2 k}{x} - 2 x^{2}$
Standard Linear Form: $\frac{d k}{d x} - \frac{2}{x} k = - 2 x^{2}$

Identifying $P (x)$ and $Q (x)$

Comparing with $\frac{d k}{d x} + P (x) k = Q (x)$ :
Found $P (x) = - \frac{2}{x}$
Found $Q (x) = - 2 x^{2}$

Calculating the Integrating Factor

Integrating Factor $I . F . = e^{\int P (x) d x}$
$I . F . = e^{\int - \frac{2}{x} d x} = e^{- 2 l n x}$
$I . F . = e^{l n (x^{- 2})} = \frac{1}{x ^{2}}$

Applying the Solution Formula

General solution: $k \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + c$
Substituting values: $k \cdot \frac{1}{x ^{2}} = \int (- 2 x^{2}) \cdot \frac{1}{x ^{2}} d x + c$

Integrating the Right Side

Simplifying the integral: $\frac{k}{x ^{2}} = \int - 2 d x + c$
Evaluating: $\frac{k}{x ^{2}} = - 2 x + c$

Re-substituting $y^{2}$

Substitute $k = y^{2}$ back: $\frac{y ^{2}}{x ^{2}} = - 2 x + c$
Multiplying by $x^{2}$ : $y^{2} = - 2 x^{3} + c x^{2}$

Final Rearrangement and Conclusion

Rearranging: $y^{2} + 2 x^{3} - c x^{2} = 0$
Since $c$ is an arbitrary constant, we can replace $- c$ with $c$ .
Final Solution: $y^{2} + 2 x^{3} + c x^{2} = 0$
Correct Option: 1

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is

(y^{2} - x^{3}) d x - x y d y = 0

. This structure suggests a non-linear form that requires a strategic transformation to solve.

The Rearrangement

Our first instinct is to isolate the derivative. By moving the

x y d y

term to the right, we obtain:

x y \frac{d y}{d x} = y^{2} - x^{3}

Dividing both sides by

x y

, we arrive at the starting point for our transformation:

\frac{d y}{d x} = \frac{y ^{2} - x ^{3}}{x y}

The Power of Substitution

We observe a

y^{2}

term in the numerator and a

y

in the denominator, which is a classic Bernoulli-style structure. Let us define a new variable

k = y^{2}

Differentiating with respect to

x

using the chain rule, we get:

\frac{d k}{d x} = 2 y \frac{d y}{d x} \Rightarrow y \frac{d y}{d x} = \frac{1}{2} \frac{d k}{d x}

Substituting this into our rearranged equation, the

y

terms simplify, leaving us with:

x (\frac{1}{2} \frac{d k}{d x}) = k - x^{3}

The Linear Path

We now have a linear differential equation in terms of

k

and

x

. Multiplying by

\frac{2}{x}

, we obtain:

\frac{d k}{d x} = \frac{2 k}{x} - 2 x^{2}

Rearranging this into the standard form

\frac{d k}{d x} + P (x) k = Q (x)

, we have:

\frac{d k}{d x} - \frac{2}{x} k = - 2 x^{2}

Here, we identify

P (x) = - \frac{2}{x}

and

Q (x) = - 2 x^{2}

. The Integrating Factor (I.F.) is calculated as follows:

I . F . = e^{\int P (x) d x} = e^{\int - \frac{2}{x} d x} = e^{- 2 l n x} = e^{l n (x^{- 2})} = \frac{1}{x ^{2}}

Final Calculation

Multiplying the linear equation by the I.F., we get:

\frac{k}{x ^{2}} = \int (- 2 x^{2}) \cdot \frac{1}{x ^{2}} d x + c

The

x^{2}

terms cancel out, simplifying the integral to:

\frac{k}{x ^{2}} = \int - 2 d x + c \Rightarrow \frac{k}{x ^{2}} = - 2 x + c

Substituting

k = y^{2}

back into the equation, we find:

\frac{y ^{2}}{x ^{2}} = - 2 x + c

Multiplying through by

x^{2}

, we obtain the final solution:

$y^{2} = - 2 x^{3} + c x^{2}$ or $y^{2} + 2 x^{3} + c x^{2} = 0$

Common Traps & FAQs

Wait, why do we reject the negative root when we substitute

y^{2} = k

That is a fantastic observation! In this specific problem, we are looking for a general solution that represents a family of curves. By substituting

y^{2} = k

, we are essentially transforming the entire equation into a new coordinate system where the relationship between

k

and

x

is linear. Since

y^{2}

is always non-negative (for real

y

k

must also be non-negative. The constant

c

in our final solution

y^{2} + 2 x^{3} + c x^{2} = 0

naturally accounts for the different branches of the curves, so we do not need to worry about the sign of

y

during the substitution process.

Why can't we just use variable separation right from the start?

I understand the temptation! Variable separation is the first tool we learn, but it only works when you can write the equation as

f (x) d x = g (y) d y

. In our equation

(y^{2} - x^{3}) d x - x y d y = 0

, the terms are inextricably linked. If you try to move all

y

terms to one side and

x

terms to the other, you will find that the

y^{2}

and

x^{3}

terms prevent a clean separation. That is exactly why we need the substitution method—it is the bridge that allows us to bypass the limitation of variable separation.

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

Great question! You should look for the standard linear form:

\frac{d k}{d x} + P (x) k = Q (x)

. Whenever you encounter a differential equation where the dependent variable (or a function of it, like

y^{2}

) appears linearly alongside its derivative, and the remaining terms are functions of

x

alone, that is your cue. The Integrating Factor,

e^{\int P (x) d x}

, is your most powerful weapon to collapse that equation into a simple, integrable form. Always check if your equation can be massaged into this structure first!

Animated Solution for Mathematics - Differential Equations: The general solution of the differential equation $(y^{2} - x^{3}) d x - x y d y = 0$ ( $x \neq = 0$ ) is :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging for $\frac{d y}{d x}$

Identifying the Substitution

Differentiating the Substitution

Substituting back into the Equation

Transforming to Linear Form

Identifying $P (x)$ and $Q (x)$

Calculating the Integrating Factor

Applying the Solution Formula

Integrating the Right Side

Re-substituting $y^{2}$

Final Rearrangement and Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Rearrangement

The Power of Substitution

The Linear Path

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The general solution of the differential equation (y2−x3)dx−xydy=0 (x=0) is :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Identifying the Substitution

Differentiating the Substitution

Substituting back into the Equation

Transforming to Linear Form

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Calculating the Integrating Factor

Applying the Solution Formula

Integrating the Right Side

Re-substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2

Final Rearrangement and Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Rearrangement

The Power of Substitution

The Linear Path

Final Calculation

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Animated Solution for Mathematics - Differential Equations: The general solution of the differential equation $(y^{2} - x^{3}) d x - x y d y = 0$ ( $x \neq = 0$ ) is :

Rearranging for $\frac{d y}{d x}$

Identifying $P (x)$ and $Q (x)$

Re-substituting $y^{2}$

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is

Let $α x = exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = lo g_{e} 2$ . Then $α + β - γ$ equals :

Let $y = y (x)$ be the solution of the differential equation $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

Let $y = y (x)$ be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Solution of the differential equation $cos x d y = y (sin x - y) d x, 0 < x < \frac{π}{2}$ is

If $x = x (y)$ is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then $x (e)$ is equal to :