1.

(x'- xy + y) (x + xy + y)​

Answer»

ong>Answer:

Your differential equation xy′−y=x2 can, assuming X≠0, be rewritten as

y′−

1

x

y=x

This is a first ORDER differential question of the form y′+P(x)y=Q(x). Such equations can be solved by finding an integrating factor, say μ, which when we multiply through by μ, the left-hand SIDE is an exact derivative. In the case of y′+Py=Q we have

μ=e ∫Pdx

In our case, P=−

1

x

and so μ=

1

|x|

.

If x>0, then μ=

1

|x|

≡

1

x

. If x<0, then μ=

1

|x|

≡−

1

x

. Multiplying through by μ=±

1

x

gives an equation EQUIVALENT to

1

x

y′−

1

x2

y=1

The left-hand side is an derivative:

(

1

x

y)′=1

Integrating both sides gives

1

x

y=x+c

It follows that y=x2+cx, where c∈R.

NOTE:

The method of using an integrating factor can be used to SOLVE u′+

1

x

u=0. In this case P=

1

x

and so μ=x. Multiplying through gives xu′+u=0 and hence (xu)′=0. It follows that xu=c and hence u=

c

x

.


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