Math, asked by saganashrees, 5 months ago

solve (x^2-y)dx+(y^2-x)y=0 if it passes through the origin

Answers

Answered by MahiSingh070
0

Answer:

(

x

2

y

)

d

x

+

(

y

2

x

)

d

y

=

0

Rewrite:

x

2

d

x

y

d

x

+

y

2

d

y

x

d

y

=

0

Rearrange the terms:

x

2

d

x

+

y

2

d

y

(

x

d

y

+

y

d

x

)

=

0

Since we know:

(

f

.

g

)

=

f

g

+

f

g

, in this problem

f

=

x

,

g

=

y

Hence,

x

d

y

+

y

d

x

=

d

(

x

y

)

x

2

d

x

+

y

2

d

y

d

(

x

y

)

=

0

Integrating both sides:

x

2

d

x

+

y

2

d

y

d

(

x

y

)

=

C

Applying the common integrals:

x

3

3

+

y

3

3

x

y

=

C

It is given that the given equation passes through the origin. So, plug in

x

=

0

,

y

=

0

to get the value of the constant.

C

=

0

Hence,

x

3

3

+

y

3

3

x

y

=

0

Therefore, the general solution of

(

x

2

y

)

d

x

+

(

y

2

x

)

d

y

=

0

is:

x

3

3

+

y

3

3

=

x

y

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