Model Paper 7 — Applied Maths STD 12 Science — Question
CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsModel Paper 73 Marks
Question
Solve the differential equation: $x \log x \frac{d y}{d x}+ y =\frac{2}{x} \log x$
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Answer
The given differential equation is $x \log x \frac{d y}{d x}+ y =\frac{2}{x} \log x$ $\Rightarrow \frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2}{x^2}$ This is a linear differential equation of the form $\frac{d y}{d x}+ Py = Q$, where $P =\frac{1}{x \log x}$ and $Q =\frac{2}{x^2}$ $\therefore$ I.F. $=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x}=e^{\int \frac{1}{t} d t}$, where $t =\log x$ $\Rightarrow$ I.F. $= e ^{\log t }= t =\log x$ Multiplying both sides of (i) by I.F. = log x, we get $\log x \frac{d y}{d x}+\frac{1}{x} y =\frac{2}{x^2} \log x$ Integrating both sides with respect to x, we get $y \log x =\int \frac{2}{x^2} \log xdx + C \left[\right.$ Using: $y ($ I.F. $)=\int Q ($ I.F. $\left.) dx + c \right]$ $\begin{array}{l}\Rightarrow y \log x =2 \int \log x x_{I I}^{-2} d x+C \\ \Rightarrow y \log x =2\left\{\log x\left(\frac{x^{-1}}{-1}\right)-\int \frac{1}{x}\left(\frac{x^{-1}}{-1}\right) d x\right\}+ C \end{array}$ $\begin{array}{l}\Rightarrow y \log x =2\left\{-\frac{\log x}{x}+\int x^{-2} d x\right\}+ C \\ \Rightarrow y \log x =2\left\{-\frac{\log x}{x}-\frac{1}{x}\right\}+ C \\ \Rightarrow y \log x =-\frac{2}{x}(1+\log x )+ C \text {, which gives the required solution. }\end{array}$
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