Understanding Engineering Mathematics

Problem 15.8

By substitutingy=xvfind the general solution of the DE

dy

dx

=

xYy

x

We note that

dy

dx

= 1 +

y

x

=a function of

y

x

and so this equation is homogeneous. So we substitutey=xvas suggested. We have

dy

dx

=

d(xv)

dx

=v+x

dv

dx

But

dy

dx

= 1 +

y

x

= 1 +v

So

x

dv

dx

= 1

or, separating the variables and integrating

∫

dv=

∫

dx

x

+C=lnx+C

Hence
v=

y

x

=lnx+C

and therefore

y=xlnx+Cx

Remember to multiply theCbyxalso!

Exercise on 15.3

Solve the differential equations

(i) y′=sinx (ii) y′=y^2

(iii) y′=x^2 y (iv) xy′= 2 x+y

In (i), (ii), (iii) give the particular solutions satisfying the conditiony( 0 )=1. In (iv) give
the solution satisfyingy( 1 )=0.

Answer

(i) y=C−cosx,y= 2 −cosx (ii) y=−

1

x+C

,y=

1

1 −x

(iii) y=Cexp

(

x^3

3

)

,y=exp

(

x^3

3

)

(iv) y= 2 xlnx+Cx,y= 2 xlnx