Solving Linear Differential Equations of First Order
Consider the ...
Question
Consider the differential equation y2dx+(x−1y)dy=0. If y(1)=1, then x is given by :
A
4−2y−e1ye
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B
3−1y−e1ye
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C
1+1y−e1ye
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D
1−1y−e1ye
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Solution
The correct option is C1+1y−e1ye dxdy+xy2=1y3 I.F.=e∫1y2dy=e−1y So x.e−1y=∫1y3e−1ydy Let −1y=t⇒1y2dy=dt ⇒I=−∫tetdt=et−tet=e−1y+1ye−1y+c ⇒x=1+1y+c.e1y Since y(1)=1 ∴c=−1e ⇒x=1+1y−1e.e1y