Find the derivative of `y = log 3x`

Answer

`a to q; b to r;c to p; dto c`

Explanation

a) `y=e^(4x)+2e^(-x), y_(1)=4e^(4x)-2e^(-x), y_(2)=16e^(4x)+2e^(-x)`,
`y_(3) = 64e^(4x) -2e^(-x)`
Now, `y_(3)-13y_(1)=64e^(4x-2e^(-x))-13(4e^(4x)-2e^(-x))`
`=12e^(4x)+24e^(-x)`
`y_(3)-13y=12(e^(4x)+2e^(-x))=12y`
`therefore K=12` and `K//3=4`
`y(dy)/(dx)=x`
`rArr ydy=xdx`
Integrating, we get
`y^(2)=x^(2)+c`
`rArr f(1)=3 rArr c=8`
`rArr y^(2)=x^(2)+8`
`f(x) =sqrt(x^(2)+8)`
`rArr f(4)=sqrt(16_8)=2sqrt(6)`
c) `y=u^(m)` or `(dy)/(dx) = mu^(m-1)(du)/(dx)`
Substituting the value of y and `(dy)/(dx)` in `2x^(4)y (dy)/(dx) + y^(4)=4x^(5)`,
We have `2x^(4)u^(m)mu^(m-1)(du)/(dx) + u^(4m) = 4x^(6)`
or `(du)/(dx) =(4x^(6)-u^(4m))/(2mx^(4)u^(2m-1))`
For homogenous equation, `4m=6` or `m=3/2`
d) `y=Ax^(m)+Bx^(-n)`
`therefore (dy)/(dx) =Amx^(m-1)-nBx^(-n-1)`
`therefore (d^(2)y)/(dx^(2)) = Am(m-1)x^(m-2)+m(n+1)Bx^(n-1)`
Putting these values in `x^(2)(d^(2)y)/(dx^(2))+2x(dy)/(dx)=12y`, we get
`m(m+1)Ax^(m)+n(n-1)bx^(-n)=12Ax^(m)+Bx^(-n)`
i.e., `m(m+1)=12` and `n(n-1)=12`
i.e., `m=3,-4` and `n=4,-3`