Question

The solution of $\frac{d^{3}y}{dx^{3}}$ – 8 $\frac{d^{2}y}{dx^{2}}$ = 0 satisfying

y(0) = 1/8, y₁ (0) = 0 and y₂ (0) = 1 is –

• A. y = $\frac { 1 } { 8 }$ $\left( \frac{e^{8x}}{8} - x + \frac{7}{8} \right)$
• B. y = $\frac { 1 } { 8 }$ $\left( \frac{e^{8x}}{8} + x + \frac{7}{8} \right)$
• C. y = $\frac { 1 } { 8 }$ $\left( \frac{e^{8x}}{8} + x - \frac{7}{8} \right)$
• D. None of these

Answer 1

Answer: (A)

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

Answer 2

We have $\frac{y_{3}}{y_{2}}$ = 8 ̃ ln y₂ = 8x + C₁

Putting x = 0, we have C₁ = log y₂ (0) = log 1 = 0

\ log y₂ = 8x or y₂ = e^8x

i.e. y₁ = $\frac{e^{8x}}{8}$ + C₂

Again, putting x = 0 , we have C₂ = – $\frac{1}{8}$

So, y₁ = $\frac{1}{8}$ (e^8x – 1) ̃ y = $\frac { 1 } { 8 }$ $\left( \frac{e^{8x}}{8} - x \right)$ + C₃

Putting x = 0, we have C₃ = $\frac{1}{8}$ – $\frac{1}{64}$ = $\frac{7}{64}$

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