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7.2 Some special types of DE

7.2.1 Separable type

Equations of the form

dydx=f(x)g(y)

are called separable. They are dealt with in the following way: Divide both sides by g(y), and integrate both sides with respect to x,

1g(y)dxdydx1g(y)dy==f(x)dxf(x)dx

Now do both integrals.

Example 7.3: 

Solve the DE

dxdy=2xy2

, given that y=12 when x=0.

Solution: 

Divide by y2, and obtain

1y2dxdy=2x

Now integrate both sides with respect to x

1y2dxdydx1y2dyy1y====2xdxx2+kx2+k1x2+k

This is the general solution, but we know that at x=0, y=12. Substituting this we find that 12=1k, therefore k=2 and

y=1x22=12x2

Example 7.4: 

Find the general solution of

2y(x+1)dxdy=4+y2

Solution: 

Rearrange as

dxdy=4+y22y(x+1)

So here f(x)=1(x+1), g(y)=2y4+y2. Divide by g(y),

2y4+y2dxdy=1x+1

Integrate both sides with respect to x

2y4+y2dxdydx2y4+y2dyln(4+y2)===1x+1dxln(x+1)+kln(x+1)+k

We write k=lnA, with A also arbitrary, but positive. We find

ln(4+y2)=ln(x+1)+lnA=ln(A(x+1))

Thus 4+y2=A(x+1), or isolating y,

y=±A(x+1)4 

Example 7.5: 

N(t) satisfies the DE

dtdN=αN

Given that N=10 at t=0 find N at t=3.

Solution: 

Here f(t)=α, i.e., a constant, and g(N)=N, so

1NdNdt1NdtdNdt1NdNlnNNN======ααdtαt+kαt+kekeαtAeαt

Since at t=0, N=10, we have A=10, and

N=10eαt

At t=3, N=10e3α.

7.2.2 linear type

These have form,

dxdy+p(x)y=q(x)
(7.1)

Method as follows

N.B. Remember the method not the final formula!

Example 7.6: 

Find the general solution of

dxdy+(tanx)y=3cosx
(7.4)

Solution: 

Here p=tanx so s=tanxdx=ln(secx) (no constant of integration needed here), es=eln(secx)=secx. Multiply both sides of (7.4) by es=secx:

secxdxdy+sec(x)tan(x)y=3cos(x)sec(x)

The l.h.s. is the differential of esy so we find

dxdsec(x)y=3

Integrate this and find

(secx)y=3x+k

Thus, finally,

y=(3x+k)cosx

Example 7.7: 

Solve the DE

xdxdy+2y=4x
(7.5)

given that y=0 when x=1.

Solution: 

Rearrange (7.5) ,

dxdy+x2y=4
(7.6)

which is of linear form with p=x2. We find

s=pdx=x2dx=2lnx=ln(x2)

and es=eln(x2)=x2. Multiply (7.6) by es=x2, and find

x2dxdy+2xy=4x2

the l.h.s. is differential of esy=x2y. Integrate this and find

x2yy==4x33+k, or4x3+kx2

This is the general solution. We know that when x=1 then y=0 so

0=43+k1

. Therefore k=43 and

y=34(x1x2)

7.2.3 Homogeneous Type

We first need to define a function of two variables:

If f(xy) is a function of 2 variables, it delivers a number on specification of x and y.
Examples:
x+y, ycos(πx), lnyx2+y2.
If x=1 and y=2 in the above we get 3, 2, 51ln2.
Now we can define a homogeneous function:
A homogeneous function of 2 variables is one where we have a sum of terms all of which have the same total power (called degree).
Examples

function degree


x2+xy+y22
x+2y1
yx2+xy21
1+yx+y2x20
xy2
x2+ynot homogeneous
x+y+1not homogeneous


There is a simple test to see if f(xy) is homogeneous. Replace x by λx and y by λy to get f(λxλy). If f(λxλy)=λnf(xy) then f is homogeneous with degree n.

Example 7.8: 

A homogeneous DE is one of type dxdy=f(xy)g(xy), with g and f both homogeneous and of the same degree.

Homogeneous DEs can be made separable by the substitution y=xv. We shall demonstrate this by means of examples:

Example 7.9: 

Find general solution of

dxdy=2xyx+2y

Solution: 

Put y=xv(x), then dxdy=v+xdxdv, so

v+xdxdv=2xxvx+2xv=2v1+2v

therefore

xdxdvdxdv==2v1+2vv=2v1+v2x12v1+v2 

which is separable. This can be solved in the standard way,

2v1+v2dxdv2v1+v2dxdvdx2v(1+v2)dv2(1+v2)dv212v(1+v2)dv2tan1v21ln(1+v2)=====x1x1dxlnx+klnx+klnx+k

And we conclude that

2tan1xy21ln1+xy=lnx+k 

(We can also replace k with lnA.)

Often we need to rearrange the equation first to get a homogeneous form, as in the following example.

Example 7.10: 

Solve

xydxdyy2=3x2

given y=1 when x=1.

Solution: 

Rearrange as

dxdy=xy3x2+y2

This is therefore a homogeneous DE. We substitute y=xv,

v+xdxdv=x2v3x2+x2v2=v3+v2=v3+v

We can now turn the crank,

xdxdvvdxdvdx2v212y2x2y2=====v3x3dx3lnx+k3lnx+k2x2(3lnx+k)

which is the general solution. Imposing the condition that for x=1, y=1, we obtain 1=2(0+k), and therefore k=12. The solution is thus

y2=2x2(2lnx+12)