Q10. Solve the differential equation
Given,
yexydx=(exy+y2)dy yexydxdy=exy+y2 ⟹exy[ydxdy−x]=y2 ⟹exy[ydxdy−x]y2=1 Let exy=t
Differentiating it w.r.t. y, we get,
ddyexy=dtdy ⟹exy⋅ddy(xy)=dtdy ⟹exy[ydxdy−x]y2=dtdy Thus from these two equations, we get, dtdy=1
⟹∫dt=∫dy ⟹t=y+C ⟹exy=y+C
CBSE Supplementary Exam 2025: Registrations open for Class 10, 12 private students today
CBSE opens application to access 10th answer scripts; verification, revaluation for Class 12 starts on May 31
CBSE supplementary exams 2025 from July 15; form submission starts on May 30
Create Your Account
To keep connected with us please login with your personal information by phone
Dont't have an account? Register Now