Solution of the differential equation x2dydx+y=1 is:
(where c is integration constant)
A
y=1+ce1x
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B
y=1+ce−1x
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C
x=1+ce1y
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D
x=1+ce−1y
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Solution
The correct option is Ay=1+ce1x The given differential equation can be written as dydx+1x2y=1x2, which is linear
On comparing with dydx+Py=Q,
We get P=1x2 and Q=1x2 I.F.=e∫(1x2)dx=e−1x
Therefore the solution is ye−1x=∫e−1x(1x2)dx+c =e−1x+c ⇒y=1+ce1x