Question

Please answer this if u know ​

Answer 1

Step-by-step explanation:

We have, y=e

x

(Acosx+Bsinx) ……(i)

differentiating with respect to x

⇒

dx

dy

=e

x

(Acosx+Bcosx)+e

x

(−Asinx+Bcosx)

⇒

dx

dy

=y+e

x

(−Asinx+Bcosx) …….(ii)

differentiating with respect to x

⇒

dx

2

d

2

y

=

dx

dy

+e

x

(−Asinx+Bcosx)+e

x

(−Acosx−Bsinx)

⇒

dx

2

d

2

y

=

dx

dy

+e

x

(−Asinx+Bcosx)−y …….(iii)

Subtracting (ii) from (iii), we get

dx

2

d

2

y

−

dx

dy

=

dx

dy

−2y

⇒

dx

2

d

2

y

−2

dx

dy

+2y=0.

This is the given differential equation. So, y=e

x

(Acosx+Bsinx) satisfies the differential equation. Hence, it is the solution of the given differential equation.

Answer 2

Answer:

I just get of 3 not of 4

Hope it helps