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

$\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)$

Let $v=x+y$, so that $\dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}$:

$\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1$

Now the ODE is separable, and we have

$\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx$

For the integral on the left, rewrite the integrand as

$\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v$

Then

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C$

and so

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

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

Given that $y(0)=\dfrac\pi2$, we find

$\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1$

so that the particular solution to this IVP is

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