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Solution of the differential equation $\frac{dy}{dx}+\frac{3x}{2y}=0$ with y(1)= 1 is : |
$2y^2+3x^2=5$ $3x^2-2y^2=5$ $2x^2+3y^2-5=0$ $2x^2+3y^2+5=0$ |
$2y^2+3x^2=5$ |
The correct answer is Option (1) → $2y^2+3x^2=5$ $\frac{dy}{dx}+\frac{3x}{2y}⇒∫2ydy=∫-3xdx$ so $y^2=-\frac{3}{2}x^2+C$ at $x = 1,y=1$ $⇒1=-\frac{3}{2}+C⇒C=\frac{5}{2}$ so eq. ⇒ $y^2=-\frac{3}{2}x^2+\frac{5}{2}$ $⇒2y^2+3x^2=5$ |
