If the order of a differential equation frac{ | vedclass

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(C) The order of the differential equation $\frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)^3+\sin \left(\frac{d y}{d x}\right)+y=0$ is $l=2$. To fi

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$\left(2 x^2-x\right) y^{\prime \prime}-\left(8 x^2-1\right) y^{\prime}+(16 x-4) y=0$

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(C) The order of the differential equation $\frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}ight)^3+\sin \left(\frac{d y}{d x}ight)+y=0$ is $l=2$. To find the degree of $\left(1+\frac{d^2 y}{d x^2}ight)^{\frac{2}{3}}=\left[2-\left(\frac{d y}{d x}ight)^3ight]^{\frac{3}{2}}$, we cube both sides: $\left(1+\frac{d^2 y}{d x^2}ight)^2 = \left[2-\left(\frac{d y}{d x}ight)^3ight]^{\frac{9}{2}}$. Squaring again to remove fractional powers, the highest derivative $\frac{d^2 y}{d x^2}$ will have power $2 	imes 2 = 4$. Thus, $m=4$. Given $y=A x^2+B e^{4 x}$. $y^{\prime}=2 A x+4 B e^{4 x}$ $y^{\prime \prime}=2 A+16 B e^{4 x}$ Eliminating $A$ and $B$: $y^{\prime \prime}-4 y^{\prime} = 2A + 16Be^{4x} - 8Ax - 16Be^{4x} = 2A - 8Ax$. $y^{\prime}-2Ax = 4Be^{4x} \implies y^{\prime \prime}-4y^{\prime} = 2A - 8Ax$. Solving for constants leads to the differential equation $\left(2 x^2-xight) y^{\prime \prime}-\left(8 x^2-1ight) y^{\prime}+(16 x-4) y=0$.

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