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Question

# Solution of the differential equation (x2+1)y′+2xy=4x2 is

A
y(1+x2)=4x33+C
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B
y(1x2)=x3+C
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C
y(1x2)=x32+C
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D
None of these
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Solution

## The correct option is A y(1+x2)=4x33+CWe have, (x2+1)y′+2xy=4x2 ⇒(x2+1)dydx+2xy=4x2 ⇒dydx+2x(x2+1)y=4x2(x2+1) On comparing that, dydx+Py=Q Now, P=2xx2+1,Q=4x2x2+1 I.F.=e∫pdx =e∫2x1+x2dx =elog(x2+1)∴elogx=x =(x2+1) Now, y×I.F.=∫Q.IFdx+C y×(x2+1)=∫4x2(x2+1)×(x2+1)dx+C y(x2+1)=∫4x2dx+C y(x2+1)=4∫x2dx+C y(x2+1)=4x33+C y(x2+1)=4x33+C Hence, this is the answer.

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