Question

The solution of the differential equation $(y+x^{2})dx=xdy, x>0$ is a curve which passes through the point (1,0). The equation of the curve is

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

Hint:

Logic Tip: Alternatively, rewrite as $\frac{dy}{dx} - \frac{1}{x}y = x$. This is a standard linear differential equation of the form $\frac{dy}{dx} + Py = Q$. The Integrating Factor is $e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$. Multiplying through yields $d(\frac{y}{x}) = dx$.

Answer 1

The Correct Option is B

Concept:
This is a first-order linear differential equation that can be rearranged into exact differential form. Specifically, recognizing the quotient rule differential $d\left(\frac{y}{x}\right) = \frac{xdy - ydx}{x^2}$ makes integration straightforward.
Step 1: Rearrange the terms.
Given: $(y + x^2)dx = xdy$ Expand the left side: $$ydx + x^2dx = xdy$$ Move all terms involving $x$ and $y$ differentials to one side to isolate $x^2dx$: $$xdy - ydx = x^2dx$$
Step 2: Create an exact differential.
Divide the entire equation by $x^2$ (valid since $x>0$): $$\frac{xdy - ydx}{x^2} = \frac{x^2dx}{x^2}$$ $$\frac{xdy - ydx}{x^2} = dx$$ Recognize that the left side is the exact derivative of the quotient $\frac{y}{x}$: $$d\left(\frac{y}{x}\right) = dx$$
Step 3: Integrate both sides.
$$\int d\left(\frac{y}{x}\right) = \int dx$$ $$\frac{y}{x} = x + C$$
Step 4: Use the initial condition to find C.
The curve passes through the point $(x=1, y=0)$. Substitute these values into the integrated equation: $$\frac{0}{1} = 1 + C$$ $$0 = 1 + C \implies C = -1$$
Step 5: Write the final equation.
Substitute $C = -1$ back into the general solution: $$\frac{y}{x} = x - 1$$ Multiply by $x$ to isolate $y$: $$y = x(x - 1)$$