Question

y(x) is the solution of (dy/dx) + (2x+1/x) y = e-2x given that y(1) = 1/2 e^-2 then y(t) is??​

Answer 1

Step-by-step explanation:

2Ae2x – 2 B.e–2x and. 2. 2. d y dx. = 4Ae2x + 4Be–2x. Thus. 2. 2. d y dx. = 4y i.e.,. 2 ... dy dx. = yex and x = 0, y = e. Find the value of y when x = 1. Solution dy dx. = yex ⇒ dy y. ∫.

Answer 2

y(x) is the solution of (dy/dx) + (2x + 1)/x y = e¯(2x) also given that y(1) = 1/2 e¯²

To find : y(t) = ?

solution : here dy/dx + (2x + 1)/x y = e^-(2x)

let P(x) = (2x + 1)/x and Q(x) = e^-(2x)

now dy/dx + P(x) y = Q(x), it is in linear form.

so integrant factor, I.f = e^{∫P(x) dx}

= e^{∫(2x + 1)/x dx

= e^{ 2x + lnx}

= e^2x × e^lnx

= e^2x × x

= x e^2x

now equation is y (I.f) = ∫Q(x) (i.f) dx

⇒y (x e^2x) = ∫e^(-2x) x e^(2x) dx

⇒xy e^2x = x²/2 + C

at x = 1

⇒1 × y(1) e² = 1²/2 + C

⇒1 × 1/2 e²e¯² = 1/2 + C

⇒C = 0

so, equation would be, xy e^2x = x²/2

⇒y = x/2e^2x

⇒y(t) = t/2e^2t

Therefore y(t) = t/(2e^2t )