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  • If the solution curve of the differential equation <img alt="\left(y-2 \log _e x\right) d x+\left(x \log _e x^2\right) d y=0, x>1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28

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___________.

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & (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{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{2 \mathrm{x} \ln \mathrm{x}}=\frac{1}{\mathrm{x}} \\ & \text { I.F }=e^{\int \frac{1}{2 x \ln x} d x} \\ & =\mathrm{e}^{\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{\frac{1}{2} \ln (\ln x)} \\ & \Rightarrow \mathrm{I} \cdot \mathrm{F}=(\ln x)^{1 / 2} \\ & \therefore \mathrm{y} \sqrt{\ln \mathrm{x}}=\int \frac{\sqrt{\ln \mathrm{x}}}{\mathrm{x}} \mathrm{dx} \quad\left(\text { Let }, \ln \mathrm{x}=\mathrm{u}^2\right) \\ & =2 \int u^2 d u & \frac{1}{x} d x=2 \mathrm{udu} \\ & \end{aligned}

\begin{aligned} & (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{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{2 \mathrm{x} \ln \mathrm{x}}=\frac{1}{\mathrm{x}} \\ & \text { I.F }=e^{\int \frac{1}{2 x \ln x} d x} \\ & =\mathrm{e}^{\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{\frac{1}{2} \ln (\ln x)} \\ & \Rightarrow \mathrm{I} \cdot \mathrm{F}=(\ln x)^{1 / 2} \\ & \therefore \mathrm{y} \sqrt{\ln \mathrm{x}}=\int \frac{\sqrt{\ln \mathrm{x}}}{\mathrm{x}} \mathrm{dx} \quad\left(\text { Let }, \ln \mathrm{x}=\mathrm{u}^2\right) \\ & =2 \int u^2 d u & \frac{1}{x} d x=2 \mathrm{udu} \\ & \end{aligned}

\begin{aligned} & \mathrm{y} \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+c \leftarrow\left(e, \frac{4}{3}\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 \end{aligned}

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manish painkra

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