Question

The equation of the curve passing through the origin and satisfying $\dfrac{dy}{dx} = (x - y)^{2}$ is

• A. $e^{2x}(1 - x + y) = 1 + x - y$
• B. $e^{2x}(1 + x - y) = 1 - x + y$
• C. $e^{2x}(1 - x + y) + (1 + x - y) = 0$
• D. $e^{2x}(1 + x + y) = 1 - x + y$

Hint:

For ODEs of the form $dy/dx = f(x-y)$, substitute $t = x-y$ (so $dt/dx = 1-dy/dx$) to separate variables.

Answer 1

The Correct Option is B

1. We start with the given differential equation: \(\dfrac{dy}{dx} = (x - y)^{2}\). This is a first-order differential equation.
2. To solve this, we can use the method of separable variables. First, we rearrange the equation as: \(dy = (x - y)^{2} dx\).
3. Let's try substituting \(v = x - y\). Then, \(y = x - v\), and differentiating with respect to \(x\) gives: \(\dfrac{dy}{dx} = 1 - \dfrac{dv}{dx}\).
4. Substituting these into the original differential equation, we get: \(1 - \dfrac{dv}{dx} = v^{2}\).
5. Rearranging gives: \(\dfrac{dv}{dx} = 1 - v^{2}\).
6. This is a separable differential equation. We can rewrite it as: \(\dfrac{dv}{1 - v^{2}} = dx\).
7. We separate the variables and integrate both sides: \(\int \dfrac{dv}{1 - v^{2}} = \int dx\).
8. The left side can be integrated using partial fraction decomposition: \(\dfrac{1}{1 - v^{2}} = \dfrac{1}{2} \left( \dfrac{1}{1-v} + \dfrac{1}{1+v} \right)\). The integral of this is \(\dfrac{1}{2}(\ln|1-v| - \ln|1+v|) = \dfrac{1}{2} \ln \left|\dfrac{1-v}{1+v}\right|\).
9. Integrating both sides, we have: \(\dfrac{1}{2} \ln \left|\dfrac{1-v}{1+v}\right| = x + C\), where \(C\) is the integration constant.
10. Exponentiating both sides gives: \(\dfrac{1-v}{1+v} = e^{2x + C}\).
11. Setting \(e^{C} = A\), we rewrite it as: \(\dfrac{1-v}{1+v} = Ae^{2x}\).
12. Solving for \(v\), we get: \(1 - v = Ae^{2x}(1 + v)\).
13. Solving for \(v\), this becomes: \(1 - v = Ae^{2x} + Ae^{2x}v\).
14. Rearranging, we have: \((1 - Ae^{2x})v = Ae^{2x} - 1\).
15. Solving for \(v\) gives: \(v = \dfrac{Ae^{2x} - 1}{1 - Ae^{2x}}\).
16. Recalling that \(v = x - y\), we substitute back: \(x - y = \dfrac{Ae^{2x} - 1}{1 - Ae^{2x}}\).
17. Simplifying, it results in: \(e^{2x}(1 + x - y) = 1 - x + y\).

Therefore, the equation of the curve is \(e^{2x}(1 + x - y) = 1 - x + y\), which matches the correct answer: $e^{2x}(1 + x - y) = 1 - x + y$.