Q20 of 462 Page 22

Find the equation to the curve satisfying and passing through (1, 0).

⇒


⇒


We can see that it is a linear differential equation.


Comparing it with


P = – , Q = 1


I.F = e∫Pdx


= edx


= edx


= elog|x + 1| – log|x|


= elog


=


Solution of the given equation is given by


y × I.F = ∫Q × I.F dx + c


⇒ y × = ∫ 1 × dx + c


⇒ y × = ∫ dx + c


⇒ y × = x + logx + c ……(1)


As the equation passing through (1,0),


0 = 1 + log1 + c


⇒ c = – 1


Putting the value of c in equation (1)


∴ y × = x + logx – 1


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