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Question : 2 of 23
Marks:
+1,
-0
Solution:
‌‌−(sin‌2x)y1=0 ‌⇒∫‌=∫sin‌2x‌dx ‌⇒∫‌=∫‌‌dx ‌⇒ln‌|y1|=‌(x−‌)+C1 ‌⇒y1=e‌(x−‌)+c1 ‌∵y1(1)=5 ‌⇒C1=ln‌5−‌+‌ ⇒y1=e‌(x−‌)+ln‌5−‌+‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(1) and
‌=(cos2x)y2 ⇒‌‌∫‌=∫cos2x‌dx=∫‌‌dx ⇒ln‌|y2|=‌(x+‌)+C2 ∵y2(1)=‌ ⇒‌‌c2=−ln‌3−‌−‌ ⇒y2=e‌(x+‌)−ln‌3−‌−‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(2) and
‌=(‌)y3 ⇒∫‌=∫(‌−1)‌dx ⇒ln‌|y3|=−‌−x+C3 ∵y3(1)=‌ ⇒‌‌C3=1+ln‌3−ln‌5 ⇒y3=e−‌−x+1+ln‌3−ln‌5‌‌‌⋅⋅⋅⋅⋅⋅⋅(3) From
eq‌n ‌ (1), (2) and (3)
y1(x)y2(x)y3(x)=e−‌ ∴‌| y1(x)y2(x)y3(x)+2x |
| e3xsin‌x |
=‌‌=‌ ∵‌‌‌e‌=0‌=‌=2
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