The differential equation having y=(a+b) e^{c | vedclass

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(D) Given the general solution $y=(a+b) e^{cx+d}$.Let $A = (a+b)e^d$. Then the equation simplifies to $y = A e^{cx}$.Here,$A$ and $c$ are the only ind

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$yy^{(2)}-\left(y^{(1)}\right)^2=0$

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(D) Given the general solution $y=(a+b) e^{cx+d}$.Let $A = (a+b)e^d$. Then the equation simplifies to $y = A e^{cx}$.Here,$A$ and $c$ are the only independent arbitrary constants.Differentiating with respect to $x$:$y^{(1)} = A c e^{cx} = c y$.Differentiating again with respect to $x$:$y^{(2)} = c y^{(1)}$.From the first derivative,we have $c = \frac{y^{(1)}}{y}$.Substituting this into the second derivative equation:$y^{(2)} = \left(\frac{y^{(1)}}{y}\right) y^{(1)}$.$y y^{(2)} = (y^{(1)})^2$.$y y^{(2)} - (y^{(1)})^2 = 0$.

The differential equation having $y=(a+b) e^{cx+d}$ as its general solution,where $a, b, c, d$ are arbitrary constants,is

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