Let y = y(x) be the solution of the differential equation (x - x3)dy = (y + yx2 - 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :

Differential Equations PYQ — JEE Main 2021 · 27 Jul · Shift 2 · Q30 with Solution

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Correct answer: B

Given $(x-x^3)\,dy=(y+y x^2-3x^4)\,dx,$ with $x>2$ .

So, $\frac{dy}{dx}=\frac{y(1+x^2)-3x^4}{x-x^3}.$

Since $x-x^3=x(1-x^2)=-x(x^2-1),$ we rewrite: $\frac{dy}{dx}=\frac{y(1+x^2)-3x^4}{x(1-x^2)}.$

Bring it to linear form: $\frac{dy}{dx}-\frac{1+x^2}{x(1-x^2)}y=\frac{-3x^4}{x(1-x^2)}=\frac{-3x^3}{1-x^2}=\frac{3x^3}{x^2-1}.$

Thus, $\frac{dy}{dx}+P(x)y=Q(x),$ where $P(x)=-\frac{1+x^2}{x(1-x^2)}=\frac{1+x^2}{x(x^2-1)}.$

So the equation is $\frac{dy}{dx}+\frac{1+x^2}{x(x^2-1)}y=\frac{3x^3}{x^2-1}.$

We need $\text{I.F.}=e^{\int \frac{1+x^2}{x(x^2-1)}dx}.$

Now decompose: $\frac{1+x^2}{x(x^2-1)}=\frac{1+x^2}{x(x-1)(x+1)}.$

A standard decomposition gives $\frac{1+x^2}{x(x^2-1)}=-\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x+1}.$

Hence,

\int\left(-\frac1x+\frac1{x-1}+\frac1{x+1}\right)dx = -\ln x+\ln(x-1)+\ln(x+1).$$ Therefore, $$\text{I.F.}=e^{-\ln x+\ln(x-1)+\ln(x+1)} =\frac{(x-1)(x+1)}{x}=\frac{x^2-1}{x}.$$ Since $x>2$, no absolute value issue affects the result. --- 3. **Multiply the differential equation by the integrating factor** Multiplying by $\dfrac{x^2-1}{x}$: $$\frac{x^2-1}{x}\frac{dy}{dx}+\frac{x^2-1}{x}\cdot \frac{1+x^2}{x(x^2-1)}y =\frac{x^2-1}{x}\cdot \frac{3x^3}{x^2-1}.$$ This simplifies to $$\frac{d}{dx}\left(\frac{x^2-1}{x}y\right)=3x^2.$$ Since $$\frac{x^2-1}{x}=x-\frac1x,$$ this is consistent. --- 4. **Integrate** $$\frac{d}{dx}\left(\frac{x^2-1}{x}y\right)=3x^2$$ Integrating, $$\frac{x^2-1}{x}y=x^3+C.$$ So, $$y=\frac{x(x^3+C)}{x^2-1}.$$ --- 5. **Use the initial condition $y(3)=3$** Substitute $x=3$, $y=3$: $$3=\frac{3(27+C)}{9-1}= rac{3(27+C)}{8}.$$ So, $$24=3(27+C),$$ $$8=27+C,$$ $$C=-19.$$ Thus, $$y=\frac{x(x^3-19)}{x^2-1}.$$ --- 6. **Find $y(4)$** $$y(4)=\frac{4(64-19)}{16-1}= rac{4\cdot 45}{15}=12.$$ --- 7. **Check the options** The value is $$y(4)=12,$$ which corresponds to **Option B**.