A) \[y=\frac{c}{{{x}^{3}}}{{e}^{{{x}^{3}}}}\]
B) \[y=c{{x}^{3}}{{e}^{-{{x}^{3}}}}\]
C) \[\frac{c}{{{x}^{3}}}{{e}^{-x}}\]
D) None of these
Correct Answer: D
Solution :
| [d] The equation is |
| \[x\int_{0}^{x}{y(t)dt=(x+1)\int_{0}^{x}{ty(t)dt,}}\] ? (i) |
| Differentiating both the sides with respect to x, we get |
| \[xy(x)+\int_{0}^{x}{y(t)dt=(x+1)xy(x)+\int_{0}^{x}{ty(t)dt}}\] |
| \[\Rightarrow \int_{0}^{x}{y(t)dt={{x}^{2}}y(x)+\int_{0}^{x}{ty(t)dt}}\] ? (ii) |
| Differentiating again with respect to x, we get |
| \[y(x)=2xy(x)+{{x}^{2}}y'(x)+xy(x)\] |
| \[\Rightarrow \,\,\,{{x}^{2}}\frac{dy}{dx}=(1-3x)y\] \[[writing\,\,y(x)=y]\] |
| \[\Rightarrow \frac{dy}{y}=\left( \frac{1-3x}{{{x}^{2}}} \right)dx=\left( \frac{1}{{{x}^{2}}}-\frac{3}{x} \right)dx\] |
| Integrating we get, \[\log y=-\frac{1}{x}-3\log x+a,a\] is constant |
| \[\Rightarrow \log y+3\log x=a-\frac{1}{x}\] |
| \[\Rightarrow \log (y{{x}^{3}})=a-\frac{1}{x}\Rightarrow y{{x}^{3}}={{e}^{a-\frac{1}{x}}}=c.{{e}^{-\frac{1}{x}}}\] |
| where \[c={{e}^{a}}\therefore y=\frac{c}{{{x}^{3}}}{{e}^{-\frac{1}{x}}}\] |
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