Question

Let \( f(x, y) = 0 \) be a solution of the homogeneous differential equation \( (2x + 5y)dx - (x + 3y)dy = 0. \)
If \( f(x + \alpha, y - 3) = 0 \) is a solution of \( (2x + 5y - 1)dx + (2 - x - 3y)dy = 0, \) then the value of \( \alpha \) is .............

Hint:

When shifting coordinates in homogeneous equations, ensure that the transformed terms match the original form for invariance.

Answer 1

Correct Answer: 7

Step 1: Given equations. 
First equation: \[ (2x + 5y)dx - (x + 3y)dy = 0. \] Second equation: \[ (2x + 5y - 1)dx + (2 - x - 3y)dy = 0. \]

Step 2: Identify translation. 
Let \( X = x + \alpha, \; Y = y - 3. \) Then \( x = X - \alpha, \, y = Y + 3. \) Substitute into second equation: \[ (2(X-\alpha) + 5(Y+3) - 1)dX + (2 - (X-\alpha) - 3(Y+3))dY = 0. \] Simplify coefficients: \[ (2X + 5Y + (15 - 2\alpha - 1))dX + (-X - 3Y + (\alpha - 7))dY = 0. \] \[ (2X + 5Y + 14 - 2\alpha)dX + (-X - 3Y + \alpha - 7)dY = 0. \]

Step 3: For \( f(X, Y) = 0 \) to remain a solution, constants must vanish. \[ 14 - 2\alpha = 0, \alpha - 7 = 0. \] Both give \( \alpha = 7. \) Wait—contradiction in scaling implies check: coefficients proportion must match original ratio. Correcting by comparing linear parts: From first equation, ratio of \( dX \) and \( dY \) parts should be preserved. Comparing, \[ (2X + 5Y) \leftrightarrow (2X + 5Y + 14 - 2\alpha), \] \[ (-X - 3Y) \leftrightarrow (-X - 3Y + \alpha - 7). \] For equality, both constants must be zero: \[ 14 - 2\alpha = 0, \alpha - 7 = 0 \Rightarrow \alpha = 7. \]

Final Answer: \[ \boxed{\alpha = 7} \]