Question

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

• A. \(\tan^{-1}(x+y) = x + c\)
• B. \(\cot^{-1}(x+y) = c\)
• C. \(\tan^{-1}(x+y) =0\)
• D. \(\cot^{-1}(x+y) = x + c\)

Answer 1

The Correct Option is A

We are given: 
\[ \frac{dy}{dx} = (x + y)^2 \] Let us use substitution. Let:
\[ z = x + y \quad \Rightarrow \quad \frac{dz}{dx} = 1 + \frac{dy}{dx} \] Since \(\frac{dy}{dx} = (x + y)^2 = z^2\), we get: \[ \frac{dz}{dx} = 1 + z^2 \] Now separate the variables: \[ \frac{1}{1 + z^2} \, dz = dx \] Integrate both sides: \[ \int \frac{1}{1 + z^2} \, dz = \int dx \quad \Rightarrow \quad \tan^{-1}(z) = x + c \] Recall \(z = x + y\), so: \[ \tan^{-1}(x + y) = x + c \] ✅ Correct answer: \(\tan^{-1}(x + y) = x + c\)

Answer 2

Given the differential equation:

\[ \frac{dy}{dx} = (x+y)^2 \] 

This equation is not directly separable or linear. We can use a substitution to simplify it. Let:

\(u = x + y\)

Differentiate both sides with respect to \(x\):

\[ \frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(y) \] \[ \frac{du}{dx} = 1 + \frac{dy}{dx} \]

Rearrange to express \(\frac{dy}{dx}\) in terms of \(\frac{du}{dx}\):

\[ \mathbf{\frac{dy}{dx} = \frac{du}{dx} - 1} \]

Now substitute \(u = x+y\) and \(\frac{dy}{dx} = \frac{du}{dx} - 1\) into the original differential equation:

\[ \frac{du}{dx} - 1 = u^2 \]

Rearrange the equation to separate variables \(u\) and \(x\):

\[ \frac{du}{dx} = u^2 + 1 \]

\[ \frac{du}{u^2 + 1} = dx \]

Integrate both sides:

\[ \int \frac{1}{u^2 + 1} \, du = \int 1 \, dx \]

The integral on the left is a standard integral form:

\[ \int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) \]

The integral on the right is:

\[ \int 1 \, dx = x + c \]

where \(c\) is the constant of integration.

Equating the results of the integration:

\[ \tan^{-1}(u) = x + c \]

Finally, substitute back \(u = x + y\):

\[ \mathbf{\tan^{-1}(x+y) = x + c} \]

Comparing this with the given options, the correct option is:

\(\tan^{-1}(x+y) = x + c\)