Match the following lists:

Answer

`atoq,s; b to p;c to p;d to q,r,s`

Explanation

p) Equation of the required parabola is of the form
`y^(2)=4a(x-h)`. Differentiating , we have
`2y(dy)/(dx)=4a or y(dy)/(dx)=2a`
The degree of this differential equation is 1 and the oder is 2.
q. We have `y=a(x+a)^(2)`...............(1)
`therefore (dy)/(dx) = 2a(x+a)`........................(2)
Dividing equations (1) by (2), we get
`y/((dy)/(dx))=(x+a)/2`
or `x+a=(2y)/(y^(1))` , where `y_(1) = (dy)/(dx)`
Substituting `a=(2y_(1))/(y_(1))-x` in equation (1), we get
`y=((2y)/(y_(1))-x)((2y)/(y_(1)))^(2)`
or `y_(1)^(3)y=4(2y-xy_(1))y^(2)`
Clearly, it is a differential equation of degree 3.
r) The given equation is `(1+3(dy)/(dx))^(2//3)=4(d^(3)y)/(dx^(3))`
Cubing, we get `(1+(dy)/(dx))^(2)=64(d^(3)y)/(dx^(2))^(3)`
Hence, order = degree=3
s) We have `y^(2)=2c(x+sqrt(c))` .............(1)
Differentiating w.r.t.x, we get `2y(dy)/(dx)=2c`
or `c=y(dy)/(dt)`
Putting in equation (1), we get
`y^(2)=2(y(dy)/(dx))x+2(y(dy)/(dx))^(3//2)`
or `(y^(2)-2xy(dy)/(dx))^(2)=4y^(3)((dy)/(dx))^(3)`
Its order is 1 and degree is 3.