CBSE Class 12 Mathematics 2026 Main Paper 65/3/1 · Q293 Markshomogeneous_equations
CBSE Class 12 Mathematics 2026 Main Paper 65/3/1 · Question 29 (Differential Equations)
Solve \(2x^2y\prime=y^2+2xy\). OR Solve \((x+1)y\prime=2e^{-y}-1\) with \(y(0)=0\).
Alternative 1 3 Marks
Solve \(2x^2y\prime=y^2+2xy\).
Answer: −x/y=(1/2)ln|x|+C on nonzero branches. Also y=0.
- For x≠0, substitute y=xv, giving y′=v+xv′ and then 2xv′=v².
- For v≠0 integrate dv/v²=dx/(2x) to obtain −1/v=(1/2)ln|x|+C.
- Replace v by y/x. The omitted v=0 branch gives y=0.
Alternative 2 3 Marks
Solve \((x+1)y\prime=2e^{-y}-1\) with \(y(0)=0\).
Answer: (x+1)(2−eʸ)=1. Equivalently y=ln(2−1/(x+1)) on x>−1/2.
- Set u=eʸ, so u′=eʸy′ and (x+1)u′=2−u.
- Then the derivative of (x+1)(2−u) is zero. Since u(0)=1, the constant is 1.
- Solve for u and take logs on the connected positive branch containing x=0.