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Question

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

A
y=18(e8x8x+78)
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B
y=18(e8x8+x+78)
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C
y=18(e8x8+x78)
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D
y=14(e8x8x+78)
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Solution

The correct option is A y=18(e8x8−x+78)Given, d3ydx3−8d2ydx2=0 ⇒y3(x)y2(x)=8 Integrating, we get, lny2(x)=8x+C1 Putting x=0, we have C1=lny2(0)=ln1=0 ∴lny2(x)=8x or y2(x)=e8x ⇒y1(x)=e8x8+C2 Again, putting x=0, we have C2=−18 So, y1(x)=18(e8x−1)⇒y=18(e8x8−x)+C3 Putting x=0, we have C3=18−164=764 Thus y=18(e8x8−x+78)

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