If y=f(x) is the solution of the differential | vedclass

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(A) Given the differential equation: $x \frac{dy}{dx} = x^2 + 3y$ Divide by $x$ $(x > 0)$: $\frac{dy}{dx} - \frac{3}{x}y = x$ This is a linear differe

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) Given the differential equation: $x \frac{dy}{dx} = x^2 + 3y$ Divide by $x$ $(x > 0)$: $\frac{dy}{dx} - \frac{3}{x}y = x$ This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\frac{3}{x}$ and $Q(x) = x$. The Integrating Factor $(IF)$ is: $IF = e^{\int P(x) dx} = e^{\int -\frac{3}{x} dx} = e^{-3 \ln x} = e^{\ln x^{-3}} = x^{-3} = \frac{1}{x^3}$. The general solution is: $y \cdot IF = \int Q(x) \cdot IF dx + C$ $y \cdot \frac{1}{x^3} = \int x \cdot \frac{1}{x^3} dx + C$ $\frac{y}{x^3} = \int x^{-2} dx + C = -x^{-1} + C = -\frac{1}{x} + C$ Given $y(2) = 4$, substitute $x = 2$ and $y = 4$: $\frac{4}{2^3} = -\frac{1}{2} + C \Rightarrow \frac{4}{8} = -\frac{1}{2} + C \Rightarrow \frac{1}{2} = -\frac{1}{2} + C \Rightarrow C = 1$. Thus, the particular solution is: $\frac{y}{x^3} = -\frac{1}{x} + 1 \Rightarrow y = x^3 - x^2$. To find $f(4)$, substitute $x = 4$: $f(4) = 4^3 - 4^2 = 64 - 16 = 48$.

If $y=f(x)$ is the solution of the differential equation $x \frac{dy}{dx} = x^2 + 3y$, $x > 0$, $y(2) = 4$, then $f(4) = $ ?

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