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

Question

asked Dec 26, 2020 18.5k views

1 Answer

Kaleem Shoukat · Answer 1 · 2020-12-30T19:30:25+0000

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

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

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