Question #79726

determine the orthogonal trajectory of families. (illustrate)

1. x^2+y^2+2ay-1=0
2. y=a(e^(-2x))

Expert's answer

Answer on Question #79726 - Math - Differential Equations

Determine the orthogonal trajectory of families. (illustrate)

Question

1.


x2+y2+2ay1=0x ^ {2} + y ^ {2} + 2 a y - 1 = 0

Solution

Differentiate respect to xx:


2x+2yy+2ay=02 x + 2 y y ^ {\prime} + 2 a y ^ {\prime} = 0x+y(y+a)=0x + y ^ {\prime} (y + a) = 0


Substitute yy^\prime to (1/y)(-1 / y^{\prime})

x(y+a)y=0x - \frac {(y + a)}{y ^ {\prime}} = 0y=(y+a)xy ^ {\prime} = \frac {(y + a)}{x}dyy+a=dxx\frac {d y}{y + a} = \frac {d x}{x}y+a=cxy + a = c x

Answer:

y+a=cxy + a = c x


For example: if a=0,c=1a = 0, c = 1


Question

2.


y=ae2xy = a e ^ {- 2 x}


Solution


y=2ae2xy ^ {\prime} = - 2 a e ^ {- 2 x}yy=12\frac {y}{y ^ {\prime}} = - \frac {1}{2}


Substitute yy^\prime to (1/y)(-1 / y^{\prime}) ..


yy=12y y ^ {\prime} = \frac {1}{2}2ydy=dx2 y d y = d xy2=x+cy ^ {2} = x + c


Answer:


y2=x+cy ^ {2} = x + c


For example: if a=1,c=1a = 1, c = 1


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