a, b, c, d are real numbers. The general solu | vedclass

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(A) The given differential equation is $\frac{dy}{dx} = \frac{ax+b}{cy+d}$.By separating the variables,we get $(cy+d)dy = (ax+b)dx$.Integrating both s

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$a=c=0$,and $b^2+d^2 
eq 0$

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(A) The given differential equation is $\frac{dy}{dx} = \frac{ax+b}{cy+d}$.By separating the variables,we get $(cy+d)dy = (ax+b)dx$.Integrating both sides,we have $\int (cy+d)dy = \int (ax+b)dx$.This results in $\frac{cy^2}{2} + dy = \frac{ax^2}{2} + bx + K$,where $K$ is the constant of integration.For the solution to represent a family of straight lines,the terms involving $x^2$ and $y^2$ must vanish.This implies that the coefficients of $x^2$ and $y^2$ must be zero,so $a=0$ and $c=0$.Substituting $a=0$ and $c=0$ into the equation,we get $dy = bdx$,which represents a family of lines provided that $b$ and $d$ are not both zero,i.e.,$b^2+d^2 
eq 0$.

$a, b, c, d$ are real numbers. The general solution of $\frac{dy}{dx} = \frac{ax+b}{cy+d}$ represents a family of straight lines,when

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