Question

If \(u=x^3\) and \(v=y^2\) transforms the ODE: \(3x^5dx - y(y^2-x^3)dy=0\) into \(\frac{dy}{du}=\frac{\lambda u}{(u-v)}\), then what is the value of \(\lambda\)?

• A. 4
• B. 2
• C. -2
• D. -4

Hint:

Substitute \(u=x^3\), \(v=y^2\) directly into the differential equation and simplify.

Answer 1

The Correct Option is C

Step 1: Write the given ODE as \(3x^5\,dx = y(y^2-x^3)\,dy\).

Step 2: Use \(u=x^3\) and \(v=y^2\). Differentiating gives \(du=3x^2\,dx\) and \(dv=2y\,dy\), so \(dx=\dfrac{du}{3x^2}\) and \(dy=\dfrac{dv}{2y}\).

Step 3: Substitute these into the ODE: \[3x^5\cdot\frac{du}{3x^2}=y(y^2-x^3)\cdot\frac{dv}{2y}\] \[x^3\,du=\frac{y^2-x^3}{2}\,dv\]

Step 4: Replace \(x^3=u\) and \(y^2=v\): \[u\,du=\frac{v-u}{2}\,dv \quad\Rightarrow\quad 2u\,du=(v-u)\,dv\]

Step 5: Rearranging for the derivative of the new variable with respect to \(u\): \[\frac{dv}{du}=\frac{2u}{v-u}=\frac{-2u}{u-v}\]

Step 6: Comparing with the transformed relation \(\dfrac{dv}{du}=\dfrac{\lambda u}{u-v}\), we identify \(\lambda=-2\). \[\boxed{\lambda=-2}\]