Solve the following differential equation dddydx = cos(x + y) Solution: dddydx = cos(x + y) ......(1) Put □ ∴ dddvd1+dydx=dvdx ∴ dddvddydx=dvdx-1 - Mathematics and Statistics

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Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c

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Solution

`("d"y)/("d"x)` = cos(x + y)    ......(1)

Put x + y = v

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

`1/(1 + cos "v")` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int sec^2 ("v"/2)  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

`tan ((x + y)/2)` = x + c

  Is there an error in this question or solution?
Chapter 1.8: Differential Equation and Applications - Q.6

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