Question

solve the given differential equation by using an appropriate substitution. the de is of the form $\frac{dy}{dx}=f(ax + by + c)$, which is given in (5) of section 2.5. $\frac{dy}{dx}=\tan^{2}(x + y)$

Explanation:

Step1: Make a substitution

Let $u = x + y$, then $\frac{du}{dx}=1+\frac{dy}{dx}$, and $\frac{dy}{dx}=\frac{du}{dx}-1$. The given differential - equation $\frac{dy}{dx}=\tan^{2}(x + y)$ becomes $\frac{du}{dx}-1=\tan^{2}u$.

Step2: Rearrange the equation

Rearrange $\frac{du}{dx}-1=\tan^{2}u$ to get $\frac{du}{dx}=1 + \tan^{2}u$. Since $1+\tan^{2}u=\sec^{2}u$, the equation is $\frac{du}{dx}=\sec^{2}u$.

Step3: Separate variables

Separate the variables: $\frac{du}{\sec^{2}u}=dx$, which simplifies to $\cos^{2}u\ du=dx$. Recall that $\cos^{2}u=\frac{1 + \cos(2u)}{2}$, so $\frac{1+\cos(2u)}{2}du=dx$.

Step4: Integrate both sides

Integrate both sides: $\int\frac{1+\cos(2u)}{2}du=\int dx$.
The left - hand side integral: $\int\frac{1+\cos(2u)}{2}du=\frac{1}{2}\int(1+\cos(2u))du=\frac{1}{2}(u+\frac{\sin(2u)}{2})+C_1$.
The right - hand side integral: $\int dx=x + C_2$.
So, $\frac{1}{2}(u+\frac{\sin(2u)}{2})=x + C$, where $C = C_2 - C_1$.

Step5: Substitute back $u=x + y$

Substitute $u=x + y$ back into the equation: $\frac{1}{2}((x + y)+\frac{\sin(2(x + y))}{2})=x + C$.
Multiply through by 2: $(x + y)+\frac{\sin(2(x + y))}{2}=2x + 2C$.
Simplify to get $y+\frac{\sin(2(x + y))}{2}=x + C$.

Answer:

$y+\frac{\sin(2(x + y))}{2}=x + C$