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Q.Find the differential equation whose general solution is y=ax+bexy = ax + be^x, where aa and bb are arbitrary constants.

Odisha ChseOdisha CHSE +2 Science Board Exam 2024Subjective· 4mImportance★★★★★
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Differentiating twice eliminates the two arbitrary constants a,ba,b, giving the second-order differential equation (1−x)y′′+xy′−y=0(1-x)y''+xy'-y=0.

Given y=ax+bexy=ax+be^x ... (1)

Differentiate: y′=a+bexy'=a+be^x ... (2)

Differentiate again: y′′=bexy''=be^x ... (3)

From (3): bex=y′′be^x=y''. Substituting into (2): a=y′−y′′a = y'-y''.

Substitute both into (1):

y=(y′−y′′)x+y′′y = (y'-y'')x + y''

y=xy′−xy′′+y′′y = xy' - xy'' + y''

y=xy′+y′′(1−x)y = xy' + y''(1-x)

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