Question

Solve the following ordinary differential equations:

a) (dy)/(dx)+y cot x=e^(cos x)
for x=(pi)/(2), y=-2

b) (d^(2)y)/(dx^(2))+(dy)/(dx)+y=0.​

Answer 1

Answer:

The solution of differential equation is $\dfrac{e^{cos x}}{sin x}$ - $\dfrac{- 3}{sin x}$

Step-by-step explanation:

Given differential equation as :

$\dfrac{dy}{dx}$ + y cot x = $e^{cosx}$

The linear differential equation form as

$\dfrac{dy}{dx}$ + p(x) y = q(x)

Integrating equation IE = $e^{\int p(x)d x}$

Or, IE = $e^{\int cotx dx}$

or, IE = $e^{ln sinx}$

Or, IE = sin x

Now,

Solution

y . IF = ∫( q(x) . IF ) dx

i.e y . sin x = ∫( $e^{cosx}$ . sin x ) dx

Or,              Let cos x = t

So, $\dfrac{dcosx}{dx}$ = $\dfrac{dt}{dx}$

Or,  - sin x dx = dt

i.e y . sin x = ∫ $e^{t}$ dt

Or, y sin x = $e^{t}$ + c

i.e y sin x = $e^{cos x}$ + c

According to question

for x = $\dfrac{\pi }{2}$  and y = - 2

i.e ( - 2)  sin $\dfrac{\pi }{2}$  = $e^{cos\frac{\pi }{2} }$ + c

Or, - 2 × 1 = $e^{0}$ + c

or, - 2 = 1 + c

i.e  c = - 2 - 1

∴   c = - 3

So, y sin x = $e^{cos x}$ - 3

Or, y = $\dfrac{e^{cos x}}{sin x}$ - $\dfrac{- 3}{sin x}$

Hence The solution of differential equation is $\dfrac{e^{cos x}}{sin x}$ - $\dfrac{- 3}{sin x}$  Answer

Answer 2

Answer:

Step-by-step explanation: