Question
Download Solution PDFअवकल समीकरण \(x^2 \frac {d^2y}{dx^2} + 4x\frac{dy}{dx}+2y=0\) का हल ________ होगा, जहाँ c1 और c2 अचर हैं।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
सहायक समीकरण के विभिन्न मूलों के लिए अवकल समीकरण का हल (पूरक फलन) नीचे दिखाया गया है।
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सहायक समीकरण के मूल |
पूरक फलन |
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m1, m2, m3, … (वास्तविक और भिन्न मूल) |
\({C_1}{e^{{m_1}x}} + {C_2}{e^{{m_2}x}} + {C_3}{e^{{m_3}x}} + \ldots\) |
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m1, m1, m3, … (दो वास्तविक और समान मूल)
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\(\left( {{C_1} + {C_2}x} \right){e^{{m_1}x}} + {C_3}{e^{{m_3}x}} + \ldots\) |
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m1, m1, m1, m4… (तीन वास्तविक और समान मूल) |
\(\left( {{C_1} + {C_2}x + {C_3}{x^2}} \right){e^{{m_1}x}} + {C_4}{e^{{m_4}x}} + \ldots\) |
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α + i β, α – i β, m3, … (काल्पनिक मूलों का एक जोड़ा) |
\({e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {C_3}{e^{{m_3}x}} + \ldots\) |
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α ± i β, α ± i β, m5, … (समान काल्पनिक मूलों के दो जोड़े) |
\({e^{\alpha x}}\left( {\left( {{C_1} + {C_2}x} \right)\cos \beta x + \left( {{C_3} + {C_4}x} \right)\sin \beta x} \right) + {C_5}{e^{{m_5}x}} + \ldots\) |
गणना:
दिया है:
\({x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = 0\)
x = et रखने पर
⇒ t = ln x
\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} ⇒ x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)
अब, उपरोक्त अवकल समीकरण निम्न प्रकार बन जाती है
D(D - 1)y + 4 Dy + 2y = 0 ⇒ (D2 + 3D + 2)y = 0
⇒ D = -1, -2;
अब हल निम्न होगा y = C1e-t + C2e-2t
x = et के मानों को रखने पर
\(\Rightarrow y = \frac{c_1}{x} + \frac{c_2}{x^2}\)
Last updated on May 26, 2025
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