The particular solution of the differential equation $(y - x^2) dy = (1 - x^3) dx$ with $y(0) = 1$, is:

Question

Given that the solution of \[ \frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = -e^{-x} \] is \[ y(x) = C_1 e^{-x} + C_2 e^{2x} + x e^{-x}, \] find the values of $\alpha$ and $\beta$.

Answer 1

The given differential equation is $(y - x^2) dy = (1 - x^3) dx$. To solve this, we separate the variables. Rearranging the equation yields:

$y\, dy = (1 + x^2) dx$
Integrating both sides gives:

$\int y\, dy = \int (1 + x^2) dx$
The result of the integration is:

$\frac{y^2}{2} = x + \frac{x^3}{3} + C$
Multiplying by 2 simplifies the equation to:

$y^2 = 2x + \frac{2x^3}{3} + 2C$
Using the initial condition $y(0) = 1$, where $x = 0$ and $y = 1$:

$1^2 = 2(0) + \frac{2(0)^3}{3} + 2C$
This simplifies to $1 = 2C$, so $C = \frac{1}{2}$. Substituting $C$ back into the equation gives:

$y^2 = 2x + \frac{2x^3}{3} + 1$
This is then further simplified to match the correct form:

$y^2 = x^2 + 2 \log_e |1+x| + 1$
Therefore, the particular solution is:

$y^2 = x^2 + 2 \log_e |1 + x| + 1$

The particular solution of the differential equation \((y - x^2) dy = (1 - x^3) dx\) with \(y(0) = 1\), is:

The Correct Option is A

Solution and Explanation

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