CBSE Class 12 Mathematics 2026 Main Paper 65/4/3 · Q345 Markslinear_differential_equations
CBSE Class 12 Mathematics 2026 Main Paper 65/4/3 · Question 34 (Differential Equations)
Solve \(ye^y\,dx=(y^3+2xe^y)\,dy\) with y(0)=1. OR Solve (x³−3xy²)dx=(y³−3x²y)dy.
Alternative 1 5 Marks
Solve \(ye^y\,dx=(y^3+2xe^y)\,dy\) with y(0)=1.
Answer: x=y²(e⁻¹−e⁻ʸ) on the branch through (0,1).
- Treat x as a function of y near y=1. The equation is dx/dy−2x/y=y²e⁻ʸ.
- Multiply by the integrating factor 1/y². Integrating gives x/y²=−e⁻ʸ+C.
- The initial condition gives C=e⁻¹.
Alternative 2 5 Marks
Solve (x³−3xy²)dx=(y³−3x²y)dy.
Answer: (x²−y²)/(x²+y²)²=C away from the origin. The level C=0 includes y=±x.
- Bring all terms to the left and set y=vx where x≠0.
- Then x v′=(1−v⁴)/(v³−3v), so (v³−3v)/(1−v⁴) dv=dx/x.
- With w=v² the left primitive is (1/2)ln|1−w|−ln(1+w). Exponentiate, absorb the sign into C, and substitute v=y/x to get (x²−y²)/(x²+y²)²=C. The straight branches y=±x are included by C=0.