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MathematicsDifferential EquationLinear DE / Red. LDEEasy2 minPYQ_2019
MathematicsEasynumerical range

The solution of the differential equationxdydx+2y=x2, (x0)withy1=1, is

Options:

Answer:
C
Solution:

Given, differential equation is dydx+2xy=x

This is a linear differential equation of type dydx+Py=Q, where P&Q are the functions of x or constants.

Thus, P=2x & Q=x

The integrating factor I.F.=ePdx

=e2x dx=e2lnx

=elnx2=x2.

The solution of the linear differential equation is y×I.F.=Q×I.F.dx+C 

So, the solution of the given differential equation is

yx2=x·x2dx+C

x2y=x3dx+C

Using xndx=xn+1n+1, we get

x2y=x44+C

Since y1=1

1·1=14+C

C=34

x2y=x44+34

 y=x24+34x2.

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Linear DE / Red. LDE
2mℹ️ Source: PYQ_2019

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