If y=100+f(x) is the particular solution of differential equation dydx=100−y and y(0)=50, then the value of f(1) is
A
50e
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B
−50e
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C
−50e
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D
50e
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Solution
The correct option is C−50e ∫dy100−y=∫dx
⇒−ln|100−y|=x+C1,C1 is a constant. ⇒ln|100−y|=−x+C(C=−C1)
where C is a constant.
If x=0, then y=50 hence, C=ln50 x=ln50−ln|100−y| ⇒ln∣∣∣50100−y∣∣∣=x ⇒50100−y=ex ⇒100−y=50e−x ⇒y=100−50e−x ⇒f(x)=−50e−x ⇒f(1)=−50e