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Exercise 4 · Q6

Q.Find the equation of the curve passing through the point (1,−1)(1, -1) whose differential equation is xydydx=(x+2)(y+2)xy\frac{dy}{dx}=(x+2)(y+2)

Yanam CbseNCERTSubjective· 3mImportance★★★★★
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Separable equation; integrate yy+2 dy=x+2x dx\dfrac{y}{y+2}\,dy=\dfrac{x+2}{x}\,dx and use (1,−1)(1,-1) to fix C=−2C=-2, giving y−2log⁡∣y+2∣=x+2log⁡∣x∣−2y-2\log|y+2|=x+2\log|x|-2.

Split each fraction: yy+2=1−2y+2\dfrac{y}{y+2}=1-\dfrac{2}{y+2} and x+2x=1+2x\dfrac{x+2}{x}=1+\dfrac{2}{x}, then integrate.

Steps

  1. Given:

xydydx=(x+2)(y+2),curve through (1,−1).xy\frac{dy}{dx}=(x+2)(y+2),\quad\text{curve through }(1,-1).

  1. Separate variables:

yy+2 dy=x+2x dx.\frac{y}{y+2}\,dy=\frac{x+2}{x}\,dx.

  1. Split each side:

(1−2y+2)dy=(1+2x)dx.\left(1-\frac{2}{y+2}\right)dy=\left(1+\frac{2}{x}\right)dx.

  1. Integrate: y−2log⁡∣y+2∣=x+2log⁡∣x∣+C.(1)y-2\log|y+2|=x+2\log|x|+C.\qquad(1) …

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