Question

Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx = cos(x + y), y(0) = π 2

Answer 1

`(\mathrm dy)/(\mathrm dx)=\cos(x+y)`

Let
`v=x+y`, so that
`(\mathrm dv)/(\mathrm dx)-1=(\mathrm dy)/(\mathrm dx)`:

`(\mathrm dv)/(\mathrm dx)=\cos v+1`

Now the ODE is separable, and we have

`(\mathrm dv)/(1+\cos v)=\mathrm dx`

Integrating both sides gives

`\displaystyle\int(\mathrm dv)/(1+\cos v)=\int\mathrm dx`

For the integral on the left, rewrite the integrand as

`\frac1{1+\cos v}\cdot(1-\cos v)/(1-\cos v)=(1-\cos v)/(1-\cos^2v)=\csc^2v-\csc v\cot v`

Then

`\displaystyle\int(\mathrm dv)/(1+\cos v)=-\cot v+\csc v+C`

and so

`\csc v-\cot v=x+C`

`\csc(x+y)-\cot(x+y)=x+C`

Given that
`y(0)=\frac\pi2`, we find

`\csc\left(0+\frac\pi2\right)-\cot\left(0+\frac\pi2\right)=0+C\implies C=1`

so that the particular solution to this IVP is

`\csc(x+y)-\cot(x+y)=x+1`