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The general solution of the differential equation (x2−yx2)dy+(y2+x2y2)dx=0(x^2 - yx^2)dy + (y^2 + x^2y^2)dx = 0 is:

Solution

✅ Correct Option: 1

The differential equation is (x2−yx2)dy+(y2+x2y2)dx=0(x^2 - yx^2)dy + (y^2 + x^2y^2)dx = 0

Factor out common terms:

x2(1−y)dy+y2(1+x2)dx=0x^2(1 - y)dy + y^2(1 + x^2)dx = 0


Rearranging:

x2(1−y)dy=−y2(1+x2)dxx^2(1 - y)dy = -y^2(1 + x^2)dx

Dividing both sides by x2y2x^2y^2:

(1−y)dyy2=−(1+x2)dxx2\frac{(1 - y)dy}{y^2} = -\frac{(1 + x^2)dx}{x^2}


Expanding the left side:

(1−y)dyy2=dyy2−dyy\frac{(1 - y)dy}{y^2} = \frac{dy}{y^2} - \frac{dy}{y}

Expanding the right side:

−(1+x2)dxx2=−dxx2−dx-\frac{(1 + x^2)dx}{x^2} = -\frac{dx}{x^2} - dx

The equation becomes:

dyy2−dyy=−dxx2−dx\frac{dy}{y^2} - \frac{dy}{y} = -\frac{dx}{x^2} - dx


Integrating both sides:

∫dyy2−∫dyy=−∫dxx2−∫dx\int \frac{dy}{y^2} - \int \frac{dy}{y} = -\int \frac{dx}{x^2} - \int dx

−1y−log⁡e∣y∣=1x−x+c-\frac{1}{y} - \log_e|y| = \frac{1}{x} - x + c


Multiplying by −1-1:

1y+log⁡e∣y∣=−1x+x−c\frac{1}{y} + \log_e|y| = -\frac{1}{x} + x - c

Replacing −c-c with cc:

log⁡e∣y∣+1y=−1x+x+c\log_e|y| + \frac{1}{y} = -\frac{1}{x} + x + c

Rearranging:

log⁡e∣y∣+1x+1y−x=c\log_e|y| + \frac{1}{x} + \frac{1}{y} - x = c

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