NCERT Class 12 Maths · Chapter 9 Differential Equations · Exercise 9.4 Question 9

✅ Step-by-Step Solution

1. After division the slope depends only on y/x (on a fixed sign component when radicals occur). Put y=vx, so y′=v+xv′.
2. The reduced equation is x dv/dx=\(- \frac{v \left(\log{\left(v \right)} - 1\right)}{\log{\left(v \right)} - 2}\). Separate dv/(\(- \frac{v \left(\log{\left(v \right)} - 1\right)}{\log{\left(v \right)} - 2}\))=dx/x on branches where division is valid.
3. Integration and back-substitution give \(\frac{1 - \log{\left(\left|{\frac{y}{x}}\right| \right)}}{y}\) = C.
4. The real log requires y/x>0. The reduced integration gives log|(1−log v)/v|=log|x|+C; clearing it gives log(y/x)+Cy=1. The straight branch y=ex is included with C=0.
5. Differentiate the resulting relation to recover the equation on intervals where the original expressions and the required slope are defined.