Math, asked by riyamagnani06, 6 months ago

xy" + (3sin^2x)y' = 5cosxy​

Answers

Answered by itzBrainlymaster
2

Answer:

ANSWER

3

dx

d

(sin(xy))+4

dx

d

(cos(xy))=

dx

d

(5)

⇒3cos(xy)

dx

d

(xy)−4sin(xy)

dx

d

(xy)=0

⇒(3cos(xy)−4sin(xy))

dx

d

(xy)=0

⇒(3cos(xy)−4sin(xy))(x

dx

dy

+y)=0

⇒3cos(xy)−4sin(xy)=0 and x

dx

dy

+y=0

⇒3cos(xy)=4sin(xy) and x

dx

dy

=−y

⇒tanxy=

4

3

and

dx

dy

=

x

−y

dx

dy

=

x

−y

Attachments:
Answered by kulkarninishant346
0

Step-by-step explanation:

3

dx

d

(sin(xy))+4

dx

d

(cos(xy))=

dx

d

(5)

⇒3cos(xy)

dx

d

(xy)−4sin(xy)

dx

d

(xy)=0

⇒(3cos(xy)−4sin(xy))

dx

d

(xy)=0

⇒(3cos(xy)−4sin(xy))(x

dx

dy

+y)=0

⇒3cos(xy)−4sin(xy)=0 and x

dx

dy

+y=0

⇒3cos(xy)=4sin(xy) and x

dx

dy

=−y

⇒tanxy=

4

3

and

dx

dy

=

x

−y

dx

dy

=

x

−y

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