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यदि ` y=a cos (log_e x )+ b sin (log _e x ) ` तब सिद्ध कीजिए की ` x^(2) y_2 +xy_1 +y=0` |
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Answer» `y=a cos (log _e x )+ b sin (log _e x ) " "...(1)` ` rArr " "y_1 =a(d)/(dx) cos (log_e x )+ b(d)/(dx) sin (log _ex)` ` =-(asin (log _e x))/(x )+(bcos log _e nx )/(x )` ` rArr " "xy_1 =- asin (log _e x )+ bcos (log _e x )` पुनः x के सापेक्ष अवकलन करने पर ` xy_2 +y_1 =-(acos (log _e x ))/(x ) -(bsin log _e x )/(x )` `rArr x^(2)y_2 +xy_1 =-[acos (log _e x )+ b sin (log _e x )]` ` " "=-y` [`because` समी० (1 ) से] ` therefore " "x^(2)y_2 +xy_1 +y=0` |
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