Question
Download Solution PDFयदि y = mex + ne-x है, तो दिए गए फलन के लिए कौन-सा संबंध सही है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
कुछ उपयोगी सूत्र निम्न हैं:
\(\rm{ d(e^{ax})\over dx} = ae^{ax}\)
गणना:
दिया गया फलन निम्न है,
y = mex + ne-x
फलन का अवकलन करने पर हमें निम्न प्राप्त होता है,
\(\rm{ dy\over dx} = me^x-ne^{-x}\)
आगे x के संबंध में अवकलन करने पर, हमें निम्न प्राप्त होता है
\(\rm{ d^2y\over dx^2} = me^x+ne^{-x}\)
आगे x के संबंध में अवकलन करने पर, हमें निम्न प्राप्त होता है
\(\rm{ d^3y\over dx^3} = me^x-ne^{-x}\)
अब, \(\rm{ d^3y\over dx^3}+{ d^2y\over dx^2}-{ dy\over dx}-y \)
= \(\rm me^x-ne^{-x}+me^x+ne^{-x}-me^x+ne^{-x}-me^x-ne^{-x}\)
= 0
अतः \(\rm{ d^3y\over dx^3}+{ d^2y\over dx^2}-{ dy\over dx}-y =0\)
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