Question

The solution of the differential equation \(x^2\frac{dy}{dx}-xy = 1\) is...

• A. \(2xy-2cx^2-1 = 0\)
• B. \(2xy+2cx^2+1 = 0\)
• C. \(2xy-2cx^2+1 = 0\)
• D. \(2x^2y-2cx+1 = 0\)

Hint:

Divide by x^2 and use the integrating factor 1/x.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The equation is first order linear. Divide by \(x^2\) to write it as \(\dfrac{dy}{dx}+Py=Q\).

Step 2: Standard form:
\[ \frac{dy}{dx}-\frac1x\,y=\frac1{x^2} \]
So \(P=-\dfrac1x\) and \(Q=\dfrac1{x^2}\).

Step 3: Integrating factor:
\[ \text{I.F.}=e^{\int-\frac1xdx}=e^{-\ln x}=\frac1x \]

Step 4: Solve:
\[ y\cdot\frac1x=\int\frac1{x^2}\cdot\frac1x\,dx=\int x^{-3}dx=-\frac1{2x^2}+c \]

Step 5: Clear the fractions:
Multiply by \(2x^2\): \(2xy=-1+2cx^2\). So
\[ 2xy-2cx^2+1=0 \]
This is option (C). Options (A) and (B) have the wrong sign on the constant 1.

Final Answer:
The solution is 2xy - 2cx^2 + 1 = 0. \[ \boxed{2xy-2cx^2+1=0} \]