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SUPPLIMENTARY PROBLEMS 1. Write the given equation in standard form

3.16. exy'-x=y' solution exy'-x=y' =exy'-y'=x ex-1y'=x y'=xex-1 ANS

3.25 dy+dx=0 Solution. dy+dx=0 dydx+1=0 y'=-1 2. The differential equations are given in both standard and differential form. Determine whether the equation in standard form are homogeneous and/or linear, if not linear, whether they are Bernoulli: determine whether the equations in differential form as given are separable and/or exact.

3.28 y'=xy+1: xy+1dx-dy=0 Solution. y'=xy+1 = y'-xy=1 this is in the form y'+pxy=qx for px=-x , qx1 thus this is linear next xy+1dx-dy=0 is not separable si9nce the variables cannot be separeted.

To check for exactness take Mx=xy+1 and Nx=-1 for an exact equation, ∂M∂y=∂N∂x

∂M∂y=x and ∂n∂y=0 thus ∂M∂y≠∂N∂x so it is not exact

3.30 y'=x2y2: -x2dx+y2dy=0 Solution. y'=x2y2 =≫ y'+0y=x21y2

The equation is of the form y'+pxy=qxyn where px=0, qx=x2 and n=-2 so it is a Bernoulli equation

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For -x2dx+y2dy=0 it is separable since the variables are separated.

To check for exactness take Mx=-x2 and Ny=y2 thus ∂M∂y=0 and ∂N∂x=0 hence its EXACT.

3.35 y'=2xy+x: 2xye-x2+xe-x2dx-e-x2dy=0

Solution.

For y'=2xy+x =≫y'-2xy=x

This is in the form y'+pxy=qx for px=-2x and qx=x Thus it’s a linear equation

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2xye-x2+xe-x2dx-e-x2dy=0 this is not separable since the variables are not separated to check for exactness take Mx= 2xye-x2+xe-x2 and Nx= -e-x2

∂M∂y=2xe-x2 and ∂N∂x=2xe-x2 thus ∂M∂y=∂N∂x and so it is EXACT…...

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