Is this correct?
Find the equation for the family of curves orthogonal to the one-parameter family y=e^cx
ANS:
y= ecx -------------- (1)
Differentiating with respect to x,
dy/dx = cecx ------------ (2)
From (1), ln(y) = cx → c= ln(y)/x
Put c= ln(y)/x and y=ecx in (2) we get,
dy/dx = ln(y)/x * y
dy/dx = y ln(y)/x
Now replace dy/dx by -1/(dy/dx) we get,
-1/(dy/dx) = yln(y)/x
-x = yln(y) dy/dx
-xdx = yln(y)dy
xdx + yln(y)dy = 0
Integrating,
∫xdx + ∫yln(y)dy = K
x2/2 + ln(y)* ∫ydy - ∫d/dy(lny) *∫y dy = K
x2/2 + ln(y)* y2/2 - ∫1/y *y2/2 dy = K
x2/2 + ln(y)* y2/2 – ½ ∫y dy = K
x2/2 + ln(y)* y2/2 – ½ *y2/2 = K
x2/2 +1/2 * y2 ln(y) – 1/4*y2 = K
Find the equation for the family of curves orthogonal to the one-parameter family y=e^cx
ANS:
y= ecx -------------- (1)
Differentiating with respect to x,
dy/dx = cecx ------------ (2)
From (1), ln(y) = cx → c= ln(y)/x
Put c= ln(y)/x and y=ecx in (2) we get,
dy/dx = ln(y)/x * y
dy/dx = y ln(y)/x
Now replace dy/dx by -1/(dy/dx) we get,
-1/(dy/dx) = yln(y)/x
-x = yln(y) dy/dx
-xdx = yln(y)dy
xdx + yln(y)dy = 0
Integrating,
∫xdx + ∫yln(y)dy = K
x2/2 + ln(y)* ∫ydy - ∫d/dy(lny) *∫y dy = K
x2/2 + ln(y)* y2/2 - ∫1/y *y2/2 dy = K
x2/2 + ln(y)* y2/2 – ½ ∫y dy = K
x2/2 + ln(y)* y2/2 – ½ *y2/2 = K
x2/2 +1/2 * y2 ln(y) – 1/4*y2 = K