If the solution curve of the differential equation (y-2 _e x ) d … - Vidyadip

Question

If the solution curve of the differential equation $\left(y-2 \log _e x\right) d x+\left(x \log _e x^2\right) d y=0, x > 1$ passes through the points $\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$, then $\alpha$ is equal to $................$.

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Answer

b
$(y-2 \ln x) d x+(2 x \ln x) d y=0$

$d y(2 x \ln x)=[(2 \ln x)-y] d x$

$\frac{d y}{d x}=\frac{1}{x}-\frac{y}{2 x \ln x}$

$\frac{d y}{d x}+\frac{y}{2 x \ln x}=\frac{1}{x}$

$\text { I.F }=e^{\int \frac{1}{2 x \ln x} d x}$

$\quad=e^{\frac{1}{2} \int \frac{d f}{t}}=e^{\frac{1}{2} \ln (\ln x)}$

$\Rightarrow I F=(\ln x)^{1 / 2}$

$\therefore y \sqrt{\ln x}=\int \frac{\sqrt{\ln x}}{x} d x$

$=2 \int u^2 d u$

$y \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+c \leftarrow\left(e, \frac{4}{3}\right) \quad\left(\text { Let }, \ln x=u^2\right)$

$\frac{4}{3}=\frac{2}{3}+c \Rightarrow c=\frac{2}{3}$

$y \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+\frac{2}{3} \leftarrow\left(e^4, \alpha\right)$

$\alpha \cdot 2=\frac{2}{3} \times 8+\frac{2}{3}$

$\alpha=3$

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