Question

Let αx=exp(x^βy^γ) be the solution of the differential equation 2x²ydy−(1−xy²) dx = 0, x>0 , y(2)=\(\sqrt {log_e2}\)​.  Then α+β−γ equals :

• A. 0
• B. $-1$
• C. 1
• D. 3

Hint:

When solving differential equations involving substitution, always verify the boundary conditions to determine the constant of integration.

Answer 1

The Correct Option is C

The given differential equation is:
\[2x^2y \, dy - (1 - xy^2) \, dx = 0.\]
Rearranging:
\[2x^2y \, dy = (1 - xy^2) \, dx.\]
\[\frac{dy}{dx} = \frac{1 - xy^2}{2x^2y}.\]
Substitute \(y^2 = t\), so that:
\[2y \, dy = dt.\]
The equation becomes:
\[x^2 \frac{dt}{dx} = 1 - xt.\]
\[\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}.\]
This is a first-order linear differential equation with integrating factor (I.F.):
\[\text{I.F.} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x.\]
Multiplying through by the integrating factor:
\[x \frac{dt}{dx} + t = \frac{1}{x}.\]
\[\frac{d}{dx}(t \cdot x) = \frac{1}{x}.\]
Integrating both sides:
\[t \cdot x = \int \frac{1}{x} dx = \ln x + C.\]
\[t = \frac{\ln x + C}{x}.\]
Substituting back \(t = y^2\):
\[y^2 \cdot x = \ln x + C.\]
Using the condition \(y(2) = \sqrt{\ln 2}\):
\[(\ln 2) \cdot 2 = \ln 2 + C.\]
\[C = \ln 2.\]
Thus:
\[y^2 \cdot x = \ln x + \ln 2.\]
\[y^2 = \frac{\ln(2x)}{x}.\]
From \(\alpha x = \exp(x^\beta y^\gamma)\):
\[\alpha x = \exp(x \cdot y^2).\]
\[\alpha x = \exp(x \cdot \frac{\ln(2x)}{x}).\]
\[\alpha x = \exp(\ln(2x)).\]
\[\alpha x = 2x.\]
Comparing with the given form \(\alpha x = \exp(x^\beta y^\gamma)\), we find:
\[\alpha = 2, \, \beta = 1, \, \gamma = 2.\]
Step 5: Calculate \(\alpha + \beta - \gamma\):
\[\alpha + \beta - \gamma = 2 + 1 - 2 = 1.\]
Conclusion: \(\alpha + \beta - \gamma = 1\) .