If frac{dy}{dx} + frac{y}{x} = x^2,then 2y(2) | vedclass

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(B) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y

Vedclass · Accepted Answer

$\frac{15}{4}$

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(B) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = x^2$. The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$. Substituting the values,we get $y \cdot x = \int x^2 \cdot x dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$. Thus,$y = \frac{x^3}{4} + \frac{C}{x}$. Now,$y(2) = \frac{2^3}{4} + \frac{C}{2} = 2 + \frac{C}{2}$. And $y(1) = \frac{1^3}{4} + \frac{C}{1} = \frac{1}{4} + C$. Calculating $2y(2) - y(1) = 2(2 + \frac{C}{2}) - (\frac{1}{4} + C) = 4 + C - \frac{1}{4} - C = 4 - \frac{1}{4} = \frac{15}{4}$.

If $\frac{dy}{dx} + \frac{y}{x} = x^2$,then $2y(2) - y(1) =$

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