Q.

If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is αx+βy+3 loge |2x+3y–γ|=6, then α+2β+3γ is equal to __________.          [2024]


Ans.

(29)

(2x + 3y – 2)dx + (4x + 6y – 7)dy = 0

dydx=–(2x+3y–2)(4x+6y–7          ... (i)

Let 2x + 3y = t

⇒ 2+3dydx=dtdx ⇒ dydx=13[dtdx–2]

Putting in (i), we get

13[dtdx]–23=–(t–2)2t–7

⇒ 13[dtdx]=23–t–22t–7=4t–14–3t+63(2t–7)

⇒ dtdx=t–82t–7 ⇒ dx=2t–7t–8dt

⇒ ∫dx+c=∫2dt+9∫dtt–8

⇒ x+c=2t+9 ln |t–8|

⇒ 2(2x+3y)+9 ln |2x+3y–8|=x+c

which is the solution of given differential equation.

Now, y(0)=3 ⇒ 2(9)+9 ln |1|=c ⇒ c=18

Putting the value of c in (i), we get

4x+6y+9 ln |2x+3y–8|=x+18

⇒ 3x+6y+9 ln |2x+3y–8|=18

⇒ x+2y+3 ln |2x+3y–8|=6

∴ α=1. β=2, γ=8

Therefore, α+2β+3γ=1+2(2)+3(8)=1+4+24=29.