Question

xdy/dx+y=x^3y^6 plz solve it
​

Answer 1

Given, xy = 4

or, x = 4/y

Now, differentiating both sides with respect to x, we get

d/dx (x) = 4 d/dx (1/y)

or, 1 = - 4/y² dy/dx

or, 4 dy/dx + y² = 0

or, 4 dy/dx = - y²

or, dy/dx = - y²/4

Now, x (dy/dx + y²)

= x (- y²/4 + y²)

= x (3y²/4)

= xy × (3y/4)

= 4 × (3y/4), since xy = 4

= 3y

Hence, proved.

Answer 2

Answer:

Step-by-step explanation:

Answer and Explanation:

Using the results from context section, we can plug in :

P

(

x

)

=

3

x

Q

(

x

)

=

x

2

−

1

d

y

d

x

+

3

x

y

=

x

2

−

1

d

(

y

x

3

)

d

x

=

(

x

2

−

1

)

x

3

y

x

3

=

∫

x

5

−

x

3

d

x

y

x

3

=

x

6

6

−

x

4

4

+

C

y

=

x

3

6

−

x

4

+

C

x

−

3