dy/dx=(y^2-8)/(yx^2+16y)=(y^2-8)/{y(x^2+16)}={(y^2-8)/y}{1/(x^2+16)}.dydx=y2−8yx2+16y=y2−8y(x2+16)={y2−8y}{1x2+16}.
rArr y/(y^2-8)dy=dx/(x^2+16).⇒yy2−8dy=dxx2+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.!
