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The solution of (3x + 3y - 4)dy + (x + y)dx = 0 is given by: |
(x + y) + log|x + y - 4| = C 3x + 3y - 4 + log|x + 4y| = C $\frac{3}{2}(x+y)+\log|x+y-2|x=C$ $e^{1/2(x+3y)}|x+y-2|=C$ |
$e^{1/2(x+3y)}|x+y-2|=C$ |
Put (x + y) = t $⇒1+\frac{dy}{dx}=\frac{dt}{dx}⇒(\frac{dt}{dx}-1)=-\frac{t}{3t-4}$ $⇒\frac{dt}{dx}=\frac{2t-4}{3t-4}⇒\int\frac{3t-4}{2t-4}dt=\int dx⇒\frac{3}{2}(x+y)+ln|2x+2y-4|=x+c$ $⇒\frac{(x+3y)}{2}+ln|2x+2y-4|=c⇒e^{1/2(x+3y)}|x+y-2|=C$ |
