JEE Main 2024 (09 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: The solution curve, of the differential equation $2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$ , passing through the point $(0, 1)$ is a conic, whose vertex lies on the line:

Q: Wait, why can't we just use the standard linear differential equation formula $y' + Py = Q$?

That is a fantastic question! The standard linear form $y' + Py = Q$ requires the derivative $\frac{dy}{dx}$ to be of the first degree and not multiplied by $y$. In our equation, we have $2y \frac{dy}{dx}$, which makes it non-linear. Because of that $y$ term, the standard formula won't work. Separation of variables is our best friend here!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not be intimidated by the differential equation. It is not a standard linear form, but it is separable. Your first step is to isolate the derivative term

\frac{d y}{d x}

and group all

y

-terms on one side and

x

-terms on the other. Visualize this as clearing the path before you begin the integration.

Hint 2

Hint 2: The Arsenal. Once you integrate both sides, you will arrive at a quadratic equation in

y

. This is the equation of a conic section. To analyze its geometric properties, specifically the vertex, you must transform this equation into the standard form of a parabola,

(y - k)^{2} = 4 a (x - h)

, by completing the square for the

y

terms.

Hint 3

Hint 3: The Execution. After completing the square, you will have an equation in the form

(y - \frac{5}{2})^{2} = - 3 (x - \frac{3}{4})

. The vertex

(h, k)

is now clearly visible. Once you have identified the vertex, simply substitute these coordinates into the given linear options to see which one satisfies the equation.

Visualized Solution

Analyze the Differential Equation

Given equation: $2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$
Goal: Isolate the $\frac{d y}{d x}$ terms to separate variables.

Rearrange and Separate Variables

Rearranging: $(2 y - 5) \frac{d y}{d x} = - 3$
Separating variables: $(2 y - 5) d y = - 3 d x$

Set up the Integration

Integrating both sides: $\int (2 y - 5) d y = \int - 3 d x$

Perform Integration

Integrating: $y^{2} - 5 y = - 3 x + C$

Apply Initial Condition $(0, 1)$

Given point: $(0, 1)$
Substitute $x = 0, y = 1$ : $1^{2} - 5 (1) = - 3 (0) + C$

Solve for Constant $C$

$- 4 = C$
Specific Equation: $y^{2} - 5 y = - 3 x - 4$

Complete the Square

$y^{2} - 5 y + \frac{25}{4} = - 3 x - 4 + \frac{25}{4}$
$(y - \frac{5}{2})^{2} = - 3 x + \frac{9}{4}$

Identify Vertex of the Parabola

$(y - \frac{5}{2})^{2} = - 3 (x - \frac{3}{4})$
Vertex $(h, k) = (\frac{3}{4}, \frac{5}{2})$

Check the Options

Check Option 1: $2 x + 3 y = 9$
LHS: $2 (\frac{3}{4}) + 3 (\frac{5}{2}) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9$

Final Conclusion

LHS = RHS. The vertex lies on this line.
Final Answer: $2 x + 3 y = 9$

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Differential Equation

The given differential equation is:

2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}

To solve this, we first isolate the derivative term

\frac{d y}{d x}

. Rearranging the terms, we obtain:

(2 y - 5) \frac{d y}{d x} = - 3

Separation and Integration

We now separate the variables to prepare for integration:

(2 y - 5) d y = - 3 d x

Integrating both sides of the equation, we get:

\int (2 y - 5) d y = \int - 3 d x

This yields the general algebraic solution:

y^{2} - 5 y = - 3 x + C

Determining the Specific Curve

We are given that the curve passes through the point

(0, 1)

. Substituting

x = 0

and

y = 1

into our general equation:

1^{2} - 5 (1) = - 3 (0) + C

Solving this gives

C = - 4

. Thus, the specific equation of the curve is:

y^{2} - 5 y = - 3 x - 4

Geometric Transformation

To identify the geometry of this curve, we complete the square for the

y

terms. Adding

\frac{25}{4}

to both sides:

y^{2} - 5 y + \frac{25}{4} = - 3 x - 4 + \frac{25}{4}

This simplifies to:

(y - \frac{5}{2})^{2} = - 3 x + \frac{9}{4}

Factoring out

- 3

on the right side, we arrive at the standard form of a parabola:

(y - \frac{5}{2})^{2} = - 3 (x - \frac{3}{4})

Final Verification

The equation represents a parabola opening to the left with its vertex at $(\frac{3}{4}, \frac{5}{2})$ .

To verify if this vertex lies on the line

2 x + 3 y = 9

, we substitute the coordinates:

2 (\frac{3}{4}) + 3 (\frac{5}{2}) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9

The vertex satisfies the equation, confirming the geometric properties of the derived curve.

Common Traps & FAQs

Wait, why can't we just use the standard linear differential equation formula

y^{'} + P y = Q

That is a fantastic question! The standard linear form

y^{'} + P y = Q

requires the derivative

\frac{d y}{d x}

to be of the first degree and not multiplied by

y

. In our equation, we have

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

, which makes it non-linear. Because of that

y

term, the standard formula won't work. Separation of variables is our best friend here!

How do I know it's a parabola and not an ellipse or hyperbola?

Great observation! Look at the powers of the variables in your final equation:

y^{2} - 5 y = - 3 x - 4

. We have a

y^{2}

term, but only a linear

x

term. In a conic section, if only one variable is squared, it is always a parabola. If both were squared with the same sign, it could be a circle or ellipse; if they had opposite signs, it would be a hyperbola.

Is there a shortcut to finding the vertex without completing the square?

You could technically use the derivative to find the vertex (since the slope at the vertex of a parabola is zero), but completing the square is the most robust and standard method for JEE. It avoids the risk of miscalculating the derivative and directly gives you the

(h, k)

form, which is exactly what you need.

Animated Solution for Mathematics - Differential Equations: The solution curve, of the differential equation $2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$ , passing through the point $(0, 1)$ is a conic, whose vertex lies on the line:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearrange and Separate Variables

Set up the Integration

Perform Integration

Apply Initial Condition $(0, 1)$

Solve for Constant $C$

Complete the Square

Identify Vertex of the Parabola

Check the Options

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Differential Equation

Separation and Integration

Determining the Specific Curve

Geometric Transformation

Final Verification

Similar Questions

The solution of the differential equation $\frac{d y}{d x} - \frac{y + 3 x}{l o g _{e} ( y + 3 x )} + 3 = 0$ is: (where $C$ is a constant of integration.)

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

If $y = y (x)$ is the solution of the differential equation $\frac{5 + e ^{x}}{2 + y} \frac{d y}{d x} + e^{x} = 0$ satisfying $y (0) = 1$ then a value of $y (lo g_{e} 13)$ is :

Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

If the curve $y = y (x)$ represented by the solution of the differential equation $(2 x y^{2} - y) d x + x d x = 0$ , passes through the intersection of the lines, $2 x - 3 y = 1$ and $3 x + 2 y = 8$ , then $∣ y (1) ∣$ is equal to

The general solution of the differential equation $1 + x^{2} + y^{2} + x^{2} y^{2} + x y \frac{d y}{d x} = 0$ is (where $C$ is a constant of integration)

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearrange and Separate Variables

Set up the Integration

Perform Integration

Apply Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,1)

Solve for Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Complete the Square

Identify Vertex of the Parabola

Check the Options

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Differential Equation

Separation and Integration

Determining the Specific Curve

Geometric Transformation

Final Verification

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is true for y(x) that satisfies the differential equation dxdy​=xy−1+x−y;y(0)=0

Apply Initial Condition $(0, 1)$

Solve for Constant $C$

The solution of the differential equation $\frac{d y}{d x} - \frac{y + 3 x}{l o g _{e} ( y + 3 x )} + 3 = 0$ is: (where $C$ is a constant of integration.)

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

If $y = y (x)$ is the solution of the differential equation $\frac{5 + e ^{x}}{2 + y} \frac{d y}{d x} + e^{x} = 0$ satisfying $y (0) = 1$ then a value of $y (lo g_{e} 13)$ is :

Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

The general solution of the differential equation $1 + x^{2} + y^{2} + x^{2} y^{2} + x y \frac{d y}{d x} = 0$ is (where $C$ is a constant of integration)

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$