If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation, y(1 +

Q: Why did we divide by $y^2$ instead of $x^2$?

Great question! We look for the denominator that makes the numerator a perfect differential. Since we have $x dy - y dx$, dividing by $y^2$ gives us $d(x/y)$. If we had $y dx - x dy$, we would divide by $x^2$ to get $d(y/x)$. It is all about matching the pattern!

Q: How do I know when to use the exact differential trick?

Whenever you see a mix of $x, y, dx, dy$ that does not fit standard linear forms, look for patterns like $x dy + y dx$ (which is $d(xy)$) or $x dy - y dx$ (which is related to $d(x/y)$). These are the 'hidden' structures that make complex problems simple.

Q: Can I solve this using the standard linear differential equation form $\frac{dy}{dx} + Py = Q$?

You could try, but it would be much more complicated. The exact differential method is the 'elegant' path. In JEE, recognizing these shortcuts is what separates the top rankers from the rest. Always look for the shortcut first!

Question

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation, $y(1 + xy) dx = x dy$, then $f(-\frac{1}{2})$ is equal to :

Options:

(A) $\frac{2}{5}$

(B) $\frac{4}{5}$

(C) $-\frac{2}{5}$

(D) $-\frac{4}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with solving the differential equation $y(1 + xy) dx = x dy$ under the condition that the curve passes through the point $(1, -1)$. Our objective is to determine the value of $f(-\frac{1}{2})$.

First, we distribute the $y$ term into the parentheses:
$$y dx + xy^2 dx = x dy$$

This rearrangement helps us isolate the terms involving $dx$ and $dy$ to identify a potential pattern.

### The Geometric Insight

We rearrange the equation by moving $y dx$ to the right side:
$$xy^2 dx = x dy - y dx$$

Observe the right-hand side, $x dy - y dx$. This expression is the numerator of the differential of the quotient $\frac{x}{y}$. To utilize this, we divide both sides by $y^2$:
$$x dx = \frac{x dy - y dx}{y^2}$$

### The Elegant Integration

We recognize that the right-hand side is equivalent to $-d(\frac{x}{y})$. The equation now simplifies to:
$$x dx = -d\left(\frac{x}{y}\right)$$

Integrating both sides, we obtain:
$$\int x dx = -\int d\left(\frac{x}{y}\right)$$
$$\frac{x^2}{2} = -\frac{x}{y} + C$$

This represents the general family of curves satisfying the given differential equation.

### Determining the Constant

We apply the initial condition that the curve passes through $(1, -1)$. Substituting $x = 1$ and $y = -1$ into our general solution:
$$\frac{(1)^2}{2} = -\frac{1}{-1} + C$$
$$\frac{1}{2} = 1 + C$$

Solving for the constant, we find $C = -\frac{1}{2}$. Thus, the particular solution is:
$$\frac{x^2}{2} = -\frac{x}{y} - \frac{1}{2}$$

### Final Calculation

To find $f(-\frac{1}{2})$, we first express $y$ in terms of $x$:
$$\frac{x^2 + 1}{2} = -\frac{x}{y}$$
$$y = -\frac{2x}{x^2 + 1}$$

Now, we substitute $x = -\frac{1}{2}$ into the function:
$$f\left(-\frac{1}{2}\right) = -\frac{2(-\frac{1}{2})}{(-\frac{1}{2})^2 + 1}$$
$$f\left(-\frac{1}{2}\right) = \frac{1}{\frac{1}{4} + 1} = \frac{1}{\frac{5}{4}}$$

The final result is:
**$$\frac{4}{5}$$**

Animated Solution for Mathematics - Differential Equations: If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :

Select Answer:

Visualized Solution

The Given Equation

Expanding the Equation

Rearranging Terms

The Exact Differential Trick

Recognizing the Derivative

Integrating Both Sides

Solving the Integral

Finding the Constant $C$

Solving for $C$

The Particular Solution

Substituting $x = - \frac{1}{2}$

Final Calculation

Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Geometric Insight

The Elegant Integration

Determining the Constant

Final Calculation

Similar Questions

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

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

If $y = y (x)$ and $\frac{2 + s i n x}{y + 1} (\frac{d y}{d x}) = - cos x, y (0) = 1$ , then $y (\frac{π}{2})$ equals

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let $f : [0, 1] \to R$ be such that $f (x y) = f (x) . f (y)$ for all $x, y \in [0, 1]$ , and $f (0) \neq = 0$ . If $y = y (x)$ satisfies the differential equation, $\frac{d y}{d x} = f (x)$ with $y (0) = 1$ , then $y (1/4) + y (3/4)$ is equal to

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0, ∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{- 1}{2})$ is equal to :

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

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

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

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Select Answer:

Visualized Solution

The Given Equation

Expanding the Equation

Rearranging Terms

The Exact Differential Trick

Recognizing the Derivative

Integrating Both Sides

Solving the Integral

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

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

The Particular Solution

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−21​

Final Calculation

Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Geometric Insight

The Elegant Integration

Determining the Constant

Final Calculation

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Finding the Constant $C$

Solving for $C$

Substituting $x = - \frac{1}{2}$

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

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

If $y = y (x)$ and $\frac{2 + s i n x}{y + 1} (\frac{d y}{d x}) = - cos x, y (0) = 1$ , then $y (\frac{π}{2})$ equals

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0, ∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{- 1}{2})$ is equal to :

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

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

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

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