What is the solution of the differential equation y' = 1+t2-2ty+y2?
Answers
Answered by
0
y' = 1 + t² - 2ty + y²
dy/dt = 1 + t² - 2ty + y²
dy/dt = 1 + (t - y)² [ ∵(a - b)² = a² - 2ab + b² ]
Now, Let ( t - y) = v
differentiate both sides with respect to t
dt/dt - dy/dt = dv/dt
1 - Dy/dt = dv/dt
1 - dv/dt = dy/dt , put it in above equation ,
1 - dv/dt = 1 + v²
⇒-dV/v² = 0
⇒1/v + C = 0
⇒1/(t - y) + C = 0 this is the solution of given equation
dy/dt = 1 + t² - 2ty + y²
dy/dt = 1 + (t - y)² [ ∵(a - b)² = a² - 2ab + b² ]
Now, Let ( t - y) = v
differentiate both sides with respect to t
dt/dt - dy/dt = dv/dt
1 - Dy/dt = dv/dt
1 - dv/dt = dy/dt , put it in above equation ,
1 - dv/dt = 1 + v²
⇒-dV/v² = 0
⇒1/v + C = 0
⇒1/(t - y) + C = 0 this is the solution of given equation
Similar questions