#### Question

Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0.

#### Solution

`dy/dx=1/cos(x+y)`

Substitute x + y = v...(1)

Differentiating w.r.t. 'x', we get

`1+dy/dx=(dv)/dx`

`dy/dx=(dv)/dx-1`

The original differential equation becomes

`(dv)/dx-1=1/cosv`

`(dv)/dx=(1+cosv)/cosv`

`int(cosvdv)/(1+cosv)=intdx`

`int((cosv+1-1)dv)/(1+cosv)=x+C`

`int(1+cosv)/(1+cosv)dv-int1/(1+cosv)dv=x+C`

`intdv-int1/(2cos^2(v/2))dv=x+C`

`v-1/2intsec^2(v/2)dv=x+C`

`v-1/2tan(v/2)dv=x+C`

Resubstituting the value of 'v' from (1), we get

`x+y-1/2tan((x+y)/2)=x+C`

`y=1/2tan((x+y)/2)=x+C`

For x = 0 and y= 0, we get C=0

Hence, particular solution is `y=1/2tan((x+y)/2)`

Is there an error in this question or solution?

#### APPEARS IN

Solution Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0. Concept: General and Particular Solutions of a Differential Equation.