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If the family of curves $y=a e^{4 x}+b e^{-x}$, where $a, b$ are arbitrary constants represents the general solution of the differential equation $f\left(x, y \frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)=0$, then $\frac{d f}{d x}=$
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The correct answer is:
$\frac{d^3 y}{d x^3}-3 \frac{d^2 y}{d x^2}-4 \frac{d y}{d x}$
We have,
On adding Eqs. (i) and (ii), we get
$y+\frac{d y}{d x}=5 a e^{4 x}$
From Eqs. (i) and (iv), we get
From Eqs. (iii), (iv) and (v), we get
$\frac{d^2 y}{d x^2}=\frac{16}{5} y+\frac{16}{5} \frac{d y}{d x}+\frac{4}{5} y-\frac{1}{5} \frac{d y}{d x}$
On adding Eqs. (i) and (ii), we get
$y+\frac{d y}{d x}=5 a e^{4 x}$
From Eqs. (i) and (iv), we get
From Eqs. (iii), (iv) and (v), we get
$\frac{d^2 y}{d x^2}=\frac{16}{5} y+\frac{16}{5} \frac{d y}{d x}+\frac{4}{5} y-\frac{1}{5} \frac{d y}{d x}$
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