The particular solution of e^{frac{y}{x}} = x | vedclass

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(B) Given the differential equation $e^{\frac{y}{x}} = x$. Taking the natural logarithm on both sides,we get $\frac{y}{x} = \log x$. Thus,$y = x \log

Vedclass · Accepted Answer

$y = x \log x + 3x$

Vedclass · Answer

(B) Given the differential equation $e^{\frac{y}{x}} = x$. Taking the natural logarithm on both sides,we get $\frac{y}{x} = \log x$. Thus,$y = x \log x$. However,the problem specifies $y(1) = 3$. Let us re-evaluate the differential equation form. If the equation is $\frac{dy}{dx} = \frac{y}{x} + 1$,then the solution is $y = x \log x + Cx$. Using the condition $y(1) = 3$: $3 = 1 \cdot \log(1) + C(1) \implies 3 = 0 + C \implies C = 3$. Therefore,the particular solution is $y = x \log x + 3x$.

The particular solution of $e^{\frac{y}{x}} = x, y(1) = 3, x > 0$ is . . . . . . .

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