#dy/dx=(y^2-8)/(yx^2+16y)=(y^2-8)/{y(x^2+16)}={(y^2-8)/y}{1/(x^2+16)}.#
# rArr y/(y^2-8)dy=dx/(x^2+16).#
This shows that the given Diff. Eqn. is of the Separable
Variable Type.
To find its General Solution (GS), we integrate it termwise.
#:. inty/(y^2-8)dy+lnc=int1/(x^2+16)dx.#
#;. 1/2int{d/dy(y^2-8)}/(y^2-8)+lnc=int1/{x^2+4^2)dx.#
#:. 1/2ln|y^2-8|+lnc=1/4 arc tan (x/4).#
#:. 2ln|y^2-8|+4lnc=arc ran (x/4), i.e., #
# ln(y^2-8)^2+lnc^4=arc tan (x/4), or, #
# ln{c^4(y^2-8)^2}=arc tan (x/4)#
#:. c^4(y^2-8)^2=e^{arc tan (x/4)}#
Hence, the GS is, # (y^2-8)^2=ke^{arc tan (x/4)}, k=1/c^4...(1)#
To find its Particular Solution (PS), we have to use the
Initial Condition # : y(x=4)=3," i.e., when, "x=4, y=3.#
Subst.ing in #(1), (9-8)^2=ke^{arc tan (4/4)}, i.e., 1=ke^(pi/4)#
#:. k=1/e^(pi/4)=e^(-pi/4).#
Therefore, from #(1)#, we have the PS # : (y^2-8)^2=e^{arc tan(x/4)-pi/4}.#
Enjoy Maths.!
