Q. Solve the following differential equation: `sec 2x tan y dx +sec 2y tan x dy = 0`

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llltkl | College Teacher | (Level 3) Valedictorian

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Given, sec2xtanydx+sec2ytanxdy=0

Rearranging, sec2xtanydx=-sec2ytanxdy

`rArr (sec2x)/tanx dx=-(sec2y)/tany dy`


`rArr int(sec2x)/tanx dx=-int(sec2y)/tany dy`

To evaluate the integral on the L.H.S

Use, `cos2x=cos^2x-sin^2x = (cos^2x-sin^2x)/(1)`

`=(cos^2x-sin^2x)/(cos^2x+sin^2x) `
dividing all terms by `cos^2x` ,
`=((cos^2x)/(cos^2x)-(sin^2x)/(cos^2x))/((cos^2x)/(cos^2x) +(sin^2x)/(cos^2x))`

hence `sec2x=1/cos(2x)= (1+tan^2x)/(1-tan^2x)`

L.H.S integral, `I_L= int(sec2x)/tanx dx`

`=int(1+tan^2x)/((1-tan^2x)tanx)dx `

`=int(sec^2x)/((1-tan^2x)tanx) dx`

let `tan x = u ` so, `du=sec^2xdx`

`I_L= int1/((1 - u²)u) du = int1/(u (1-u)(1+u))du`

Decomposing this into partial fractions,

say, `I_L=int(A/u+B/(1-u)+C/(1+u))du`

Upon comparing the coefficients of `u^2` , `u ` and the constant term, then solving, we get A=1, B=1/2 and C=-1/2

So, `I_L=int(1/u+(1/2)/(1-u)-(1/2)/(1+u))du`

`=lnu-(1/2)ln(1-u)- (1/2)ln(1+u)`


Being u = tan x,

`I_L= ln(tanx/(1-tan^2x)^(1/2))`

Similarly, the integral on the right hand side,

`I_R=- ln(tany/(1-tan^2y)^(1/2))`
and the overall differential equation, after integrating, takes the form,

`ln(tanx/(1-tan^2x)^(1/2)) =- ln(tany/(1-tan^2y)^(1/2))+lnk ` (where, lnk is the constant of integration)

So, `ln(tanx/(1-tan^2x)^(1/2)) + ln(tany/(1-tan^2y)^(1/2))=lnk`

`rArr ln(tanx/(1-tan^2x)^(1/2)*tany/(1-tan^2y)^(1/2))=lnk`

`rArr (tanx/(1-tan^2x)^(1/2)* tany/(1-tan^2y)^(1/2))=k`

`rArr tanx/sqrt(1-tan^2x)=k*sqrt(1-tan^2y)/tany`