Question

The solution of $\frac{dy}{dx} = (x + y)^2$ is

• A. $\tan^{-1}(x + y) = x + c$, where c is the constant of integration
• B. $x + y = \tan x + c$, where c is the constant of integration
• C. $x + y = \cot^{-1} x + c$, where c is the constant of integration
• D. $x + y = \sin^{-1}(x + y) + c$, where c is the constant of integration

Hint:

Always try the substitution $v = x+y$ when they appear together inside a function in a DE.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given differential equation is of the form $\frac{dy}{dx} = f(ax + by + c)$. Such equations can be solved by substituting the linear expression $ax + by + c$ with a new variable.
Step 2: Key Formula or Approach:
Let $x + y = v$. Differentiate with respect to $x$ to find an expression for $\frac{dy}{dx}$ in terms of $\frac{dv}{dx}$, and substitute it back into the original equation to separate variables.
Step 3: Detailed Explanation:
Given differential equation:
\[ \frac{dy}{dx} = (x + y)^2 \] Substitute $v = x + y$.
Differentiating both sides with respect to $x$:
\[ \frac{dv}{dx} = 1 + \frac{dy}{dx} \] \[ \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \] Substitute this back into the original differential equation:
\[ \frac{dv}{dx} - 1 = v^2 \] \[ \frac{dv}{dx} = v^2 + 1 \] Now, separate the variables $v$ and $x$:
\[ \frac{dv}{v^2 + 1} = dx \] Integrate both sides:
\[ \int \frac{dv}{v^2 + 1} = \int 1 dx \] The integral of $\frac{1}{v^2 + 1}$ is $\tan^{-1}(v)$.
\[ \tan^{-1}(v) = x + c \] Finally, substitute back $v = x + y$:
\[ \tan^{-1}(x + y) = x + c \] Step 4: Final Answer:
The solution is $\tan^{-1}(x + y) = x + c$.