Question
Download Solution PDFजब x = 0 है तो अवकल समीकरण \(\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = {e^x},\;y = 3\) और \(\frac{{dy}}{{dx}} = 3\) का समाधान क्या है
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
nवीं कोटि के एक रैखिक अवकल समीकरण के समाधान में दो भाग शामिल हैं। वे हैं:
- पूरक फलन
- विशेष समाकल
यानी y = C.F. + P.I.
गणना:
दिया हुआ:
\(\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = {e^x},\;y = 3\;and\;\frac{{dy}}{{dx}} = 3\;when\;x=0\)
पूरक फलन (CF):
R.H.S = 0 डालें और मूलों की गणना करें।
\(\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = 0\)
मूलों को खोजने के लिए d/dx = m डालें।
∴ m2 - 3m + 2 = 0
∴ m = 1 और 2
∴ y = C1ex + C2e2x
विशेष समाकल (PI):
\(\frac{1}{D^2\;-\;3D\;+\;2}e^x\)
\(\frac{1}{2D\;-\;3}xe^x\) [∵ f(D) = 0 जब D = 1]
∴ P.I = -xex
सामान्य समाधान yy = C1ex + C2e2x - xex है
जब x = 0, y = 3।
∴ C 1 + C 2 = 3 ...... (1)
जब x = 0, dy/dx = 3।
\(\frac{dy}{dx}=C_1e^x\;+\;2C_2e^{2x}\;-\;xe^x\;-\;e^x\)
∴ 3 = C1 + 2C2 - 1
∴ C1 + 2C2 = 4 ........ (2)
(1) और (2) को हल करना C1 = 2 और C2 = 1 देता है।
∴ y = 2ex + e2x - xex
Last updated on Jun 21, 2025
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