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प्रश्न
Solve the following differential equation:
`("d"y)/("d"x) = tan^2(x + y)`
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उत्तर
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
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