Question

If x dy= y(dx + ydy), x(1)=1, y(x)>0, then y (-3) is?

Answer 1

Let's solve the given differential equation step by step:

1. Rewrite the differential equation:
The given equation is: \[ x \, dy = y \, (dx + y \, dy) \] Expanding it: \[ x \, dy = y \, dx + y^2 \, dy \] Rearranging the terms: \[ x \, dy - y \, dx = y^2 \, dy \]

2. Divide by \( x^2 \):
Dividing the entire equation by \( x^2 \): \[ \frac{x \, dy - y \, dx}{x^2} = \frac{y^2}{x^2} \, dy \]

3. Recognize the left side as the derivative of \( y/x \):
The left side can be recognized as the derivative of \( y/x \), which gives us: \[ d\left( \frac{y}{x} \right) = \frac{y^2}{x^2} \, dy \]

4. Substitute \( v = \frac{y}{x} \), so \( y = vx \):
Let \( v = \frac{y}{x} \), so \( y = vx \). Taking the derivative of \( y \) with respect to \( x \): \[ dy = v \, dx + x \, dv \] Substituting this into the equation: \[ d\left( \frac{y}{x} \right) = v^2 \, dy \] This gives: \[ dv = v^2 \left( v \, dx + x \, dv \right) \] Rearranging the terms: \[ dv - v^2 x \, dv = v^3 \, dx \] Simplifying: \[ dv \left( 1 - v^2 x \right) = v^3 \, dx \] \[ \frac{dv}{v^3} = \frac{dx}{(1 - v^2 x)} \] This approach leads to a complex differential equation.

Alternative Approach:

Let's go back to the original equation: \[ x \, dy - y \, dx = y^2 \, dy \] 5. Divide by \( y^2 \):
Dividing both sides by \( y^2 \): \[ \frac{x \, dy - y \, dx}{y^2} = dy \] This simplifies to: \[ - d\left( \frac{x}{y} \right) = dy \] 6. Integrate both sides:
Integrating both sides: \[ - \int d\left( \frac{x}{y} \right) = \int dy \] This gives: \[ - \frac{x}{y} = y + C \] Rearranging: \[ x = - y^2 - Cy \] So, we have: \[ x + y^2 + Cy = 0 \]

7. Apply the initial condition \( x(1) = 1 \):
We are given that when \( x = 1 \), \( y = 1 \). Substituting these values into the equation: \[ 1 + 1^2 + C(1) = 0 \] Simplifying: \[ 1 + 1 + C = 0 \] \[ 2 + C = 0 \] So, \( C = -2 \).

8. Substitute \( C = -2 \) back into the equation:
Substituting \( C = -2 \) into the equation: \[ x + y^2 - 2y = 0 \]

9. Solve for \( y \):
The equation becomes: \[ y^2 - 2y + x = 0 \] Using the quadratic formula to solve for \( y \): \[ y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(x)}}{2(1)} \] Simplifying: \[ y = \frac{2 \pm \sqrt{4 - 4x}}{2} \] \[ y = 1 \pm \sqrt{1 - x} \]

10. Apply the condition \( y(x) > 0 \):
Since \( y(x) > 0 \), we must choose the positive root: \[ y = 1 + \sqrt{1 - x} \]

11. Find \( y(-3) \):
To find \( y(-3) \), substitute \( x = -3 \) into the equation: \[ y(-3) = 1 + \sqrt{1 - (-3)} \] \[ y(-3) = 1 + \sqrt{4} \] \[ y(-3) = 1 + 2 \] \[ y(-3) = 3 \]

Final Answer:
Therefore, \( y(-3) = 3 \).