Question

Let $y=y(x)$ be the solution of the differential equation $(x-x^3)dy = (y+yx^2-3x^4)dx$, $x>2$. If $y(3)=3$, then $y(4)$ is equal to :

• A. 12
• B. 8
• C. 16
• D. 4

Hint:

When a differential equation seems complex, try to rearrange it into a standard form like linear ($y'+P(x)y=Q(x)$) or exact. For linear equations, the key is to correctly calculate the integrating factor, $e^{\int P(x)dx}$.

Answer 1

The Correct Option is A

The given differential equation is $(x-x^3)dy = (y+yx^2-3x^4)dx$.
Let's rearrange it to see if it fits a standard form.
$x(1-x^2)dy = (y(1+x^2) - 3x^4)dx$.
$\frac{dy}{dx} = \frac{y(1+x^2) - 3x^4}{x(1-x^2)}$.
$\frac{dy}{dx} = \frac{y(1+x^2)}{x(1-x^2)} - \frac{3x^4}{x(1-x^2)}$.
$\frac{dy}{dx} - \frac{y(1+x^2)}{x(1-x^2)} = -\frac{3x^3}{1-x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Here, $P(x) = -\frac{1+x^2}{x(1-x^2)} = \frac{1+x^2}{x(x^2-1)}$.
And $Q(x) = -\frac{3x^3}{1-x^2} = \frac{3x^3}{x^2-1}$.
Let's find the integrating factor (I.F.).
I.F. = $e^{\int P(x)dx} = e^{\int \frac{1+x^2}{x(x^2-1)}dx}$.
Using partial fractions for the integrand: $\frac{1+x^2}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.
$A(x^2-1) + Bx(x+1) + Cx(x-1) = 1+x^2$.
If $x=0, -A=1 \Rightarrow A=-1$.
If $x=1, 2B=2 \Rightarrow B=1$.
If $x=-1, 2C=2 \Rightarrow C=1$.
So, $\int P(x)dx = \int (-\frac{1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) dx = -\ln x + \ln(x-1) + \ln(x+1) = \ln\left(\frac{x^2-1}{x}\right)$.
I.F. = $e^{\ln\left(\frac{x^2-1}{x}\right)} = \frac{x^2-1}{x}$.
The solution is $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$.
$y \left(\frac{x^2-1}{x}\right) = \int \frac{3x^3}{x^2-1} \cdot \frac{x^2-1}{x} dx + C$.
$y \left(\frac{x^2-1}{x}\right) = \int 3x^2 dx + C = x^3 + C$.
We are given the condition $y(3)=3$.
$3 \left(\frac{3^2-1}{3}\right) = 3^3 + C$.
$3 \left(\frac{8}{3}\right) = 27 + C \Rightarrow 8 = 27 + C \Rightarrow C = -19$.
The solution is $y \left(\frac{x^2-1}{x}\right) = x^3 - 19$.
Now we need to find $y(4)$.
$y(4) \left(\frac{4^2-1}{4}\right) = 4^3 - 19$.
$y(4) \left(\frac{15}{4}\right) = 64 - 19 = 45$.
$y(4) = 45 \times \frac{4}{15} = 3 \times 4 = 12$.
So, $y(4) = 12$.