Question

solve the given initial - value problem. the de is of the form $\frac{dy}{dx}=f(ax + by + c)$. $\frac{dy}{dx}=cos(x + y),y(0)=\frac{pi}{2}$

Explanation:

Step1: Use substitution

Let \(u=x + y\), then \(\frac{du}{dx}=1+\frac{dy}{dx}\), and the differential - equation \(\frac{dy}{dx}=\cos(x + y)\) becomes \(\frac{du}{dx}-1=\cos u\).

Step2: Rearrange the equation

Rearrange \(\frac{du}{dx}-1=\cos u\) to \(\frac{du}{dx}=1 + \cos u\). Then, separate the variables: \(\frac{du}{1+\cos u}=dx\).

Step3: Integrate both sides

Recall the double - angle formula \(1+\cos u = 2\cos^{2}\frac{u}{2}\). So, \(\int\frac{du}{1+\cos u}=\int dx\). The left - hand side integral \(\int\frac{du}{2\cos^{2}\frac{u}{2}}=\frac{1}{2}\int\sec^{2}\frac{u}{2}du\). Let \(t=\frac{u}{2}\), \(dt=\frac{1}{2}du\), then \(\frac{1}{2}\int\sec^{2}\frac{u}{2}du=\int\sec^{2}t\ dt=\tan t+C=\tan\frac{u}{2}+C\). The right - hand side integral \(\int dx=x + C\). So, \(\tan\frac{u}{2}=x + C\).

Step4: Substitute back \(u=x + y\)

We get \(\tan\frac{x + y}{2}=x + C\).

Step5: Use the initial condition

Given \(y(0)=\frac{\pi}{2}\), substitute \(x = 0\) and \(y=\frac{\pi}{2}\) into \(\tan\frac{x + y}{2}=x + C\). Then \(\tan\frac{0+\frac{\pi}{2}}{2}=0 + C\), \(\tan\frac{\pi}{4}=C\), so \(C = 1\).

Step6: Write the final solution

The solution of the initial - value problem is \(\tan\frac{x + y}{2}=x + 1\).

Answer:

\(\tan\frac{x + y}{2}=x + 1\)