Question

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

• A. $\cos(\frac{y}{x}) = cx$
• B. $\sin(\frac{y}{x}) = cx$
• C. $\cos(\frac{y}{x}) = cy$
• D. $\sin(\frac{y}{x}) = cy$

Answer 1

Answer: (B)

$\sin(\frac{y}{x}) = cx$

Answer 2

The given differential equation is: $\frac{dy}{dx} = \tan\left(\frac{y}{x}\right) + \left(\frac{y}{x}\right)$

This is a homogeneous differential equation because it can be expressed in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$.

To solve homogeneous differential equations, we use the substitution $y = vx$. Differentiating both sides with respect to $x$, we get: $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the original differential equation: $v + x\frac{dv}{dx} = \tan\left(\frac{vx}{x}\right) + \left(\frac{vx}{x}\right)$ $v + x\frac{dv}{dx} = \tan(v) + v$

Subtract $v$ from both sides: $x\frac{dv}{dx} = \tan(v)$

This is a separable differential equation. Separate the variables $v$ and $x$: $\frac{dv}{\tan(v)} = \frac{dx}{x}$ $\cot(v) dv = \frac{dx}{x}$

Now, integrate both sides: $\int \cot(v) dv = \int \frac{dx}{x}$

The integral of $\cot(v)$ is $\ln|\sin(v)|$, and the integral of $\frac{1}{x}$ is $\ln|x|$. We add a constant of integration, which can be written as $\ln|C|$ for convenience: $\ln|\sin(v)| = \ln|x| + \ln|C|$

Using the logarithm property $\ln a + \ln b = \ln(ab)$: $\ln|\sin(v)| = \ln|Cx|$

Exponentiate both sides to remove the logarithm: $|\sin(v)| = |Cx|$

Since $C$ is an arbitrary constant, it can absorb the signs, so we can write: $\sin(v) = Cx$

Finally, substitute back $v = \frac{y}{x}$: $\sin\left(\frac{y}{x}\right) = Cx$