The solution of d3 y-d x3-8 d2 y-d x2-0 satisfying y0-1-8- y10-0 and y20-1 is -here -ynx-dp y-d xx-A- y-1-88 x-8-x-7-8B- y-1-88 x-8 x-x-7-8C- y-1-48 x-8 x-x-7-8D- y-1-88 x-8-x-7-8

The correct option is A y =1/8 ( e^8x/8 x + 7/8 )Given, d^3y/dx^3 8d^2y/dx^2=0 ⇒y^3(x)/y^2(x)=8 Integrating, we get, ln y^2(x) =8x + C_1 Putting x =0, we hav ...

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Standard XII

Mathematics

Formation of a Differential Equation from a General Solution

The solution ...

Question

The solution of $\frac{d^{3} y}{d x^{3}} - 8 \frac{d^{2} y}{d x^{2}} = 0$ satisfying $y (0) = \frac{1}{8}, y^{1} (0) = 0$ and $y^{2} (0) = 1$ is [here $y^{n} (x) = \frac{d^{n} y}{d x^{n}}]$

y = \frac{1}{8} (\frac{e^{8 x}}{8} - x + \frac{7}{8})

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y = \frac{1}{8} (\frac{e^{8 x}}{8} + x + \frac{7}{8})

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y = \frac{1}{8} (\frac{e^{8 x}}{8} + x - \frac{7}{8})

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y = \frac{1}{4} (\frac{e^{8 x}}{8} - x + \frac{7}{8})

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Solution

The correct option is A $y = \frac{1}{8} (\frac{e^{8 x}}{8} - x + \frac{7}{8})$
Given, $\frac{d^{3} y}{d x^{3}} - 8 \frac{d^{2} y}{d x^{2}} = 0$
$\Rightarrow \frac{y^{3} (x)}{y^{2} (x)} = 8$
Integrating, we get,
$ln y^{2} (x) = 8 x + C_{1}$
Putting $x = 0,$ we have $C_{1} = ln y^{2} (0) = ln 1 = 0$
$∴ ln y^{2} (x) = 8 x$ or $y^{2} (x) = e^{8 x}$
$\Rightarrow y^{1} (x) = \frac{e^{8 x}}{8} + C_{2}$
Again, putting $x = 0,$ we have $C_{2} = - \frac{1}{8}$
So, $y^{1} (x) = \frac{1}{8} (e^{8 x} - 1) \Rightarrow y = \frac{1}{8} (\frac{e^{8 x}}{8} - x) + C_{3}$
Putting $x = 0,$ we have $C_{3} = \frac{1}{8} - \frac{1}{64} = \frac{7}{64}$
Thus $y = \frac{1}{8} (\frac{e^{8 x}}{8} - x + \frac{7}{8})$

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The solution of d3ydx3−8d2ydx2=0 satisfying y(0)=18, y1(0)=0 and y2(0)=1 is [here yn(x)=dnydxn]

The solution of $\frac{d^{3} y}{d x^{3}} - 8 \frac{d^{2} y}{d x^{2}} = 0$ satisfying $y (0) = \frac{1}{8}, y^{1} (0) = 0$ and $y^{2} (0) = 1$ is [here $y^{n} (x) = \frac{d^{n} y}{d x^{n}}]$