Question

Solve the following differential equation:

`("d"y)/("d"x) = tan^2(x + y)`

Answer 1

Given `("d"y)/("d"x) = tan^2(x + y)`   .......(1)

Take x + y = t

`1 + ("d"y)/("d"x) = "dt"/("d"x)`

`("d"y)/("d"x) = "dt"/("d"x) - 1`

∴ Equation (1) can be written as

`("d"y)/("d"x) = tan^2(x + y)`

`"dt"/("d"x) - 1 = tan^2"t"`

`"dt"/("d"x) = tan^2"t" + 1`

`"dt"/("d"x) = sec^2"t"`   ........(∵ 1 +tan²θ = sec²θ)

`"dt"/(sec^2"t")` = dx

cos²t dt = dx

`((1 + cos^2"t")/2)  dt"` = dx   ......`(∵ cos^2theta = (1 + cos^2theta)/2)`

Takig integration on both sides, we get

`1/2 int(1 + cos^2"t")  "dt"= int "d"x`

`1/2["t" + (sin^2"t")/2]` = x + c

`1/2["t" + (2sin"t" cos"t")/2]` = x + c

`1/2["t" + sin"t" cos"t"]` = x + c  .......(∵ t = x + y)

`1/2[x + y + sin(x + y) cos(x + y)]` = x + c