TestHub
TestHub

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationVariable separableEasy2 minPYQ_2021
MathematicsEasynumerical range

Ifdy dx=2xy+2y·2x2x+2x+yloge2, y0=0,then fory=1,the value ofxlies in the interval :

Options:

Answer:
A
Solution:

Given differential equation is

dydx=2x·y+2y·2x2x+2x+yloge2

dydx=2xy+2y2x1+2yloge2

1+2yloge2y+2ydy=dx

dy+2yy+2y=dx

lny+2y=x+Cf'xfxdx=lnfx+C

Now y0=0

C=0

lny+2y=x

Now for y=1 we have

x=ln1+2=ln31, 2

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Variable separable
2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...