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Question Number 183209 by Mastermind last updated on 23/Dec/22

$$\mathrm{Solve}\mathrm{the}\mathrm{Differential}\mathrm{equation}\mathrm{below}$$$$\frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}+8\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+12\frac{\mathrm{dy}}{\mathrm{dx}}=0$$$$$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by aleks041103 last updated on 24/Dec/22

$$y^{\left(3\right)}+8y^{\left(2\right)}+12y^{\left(1\right)}=0$$$$z=y^{\left(1\right)}=e^{rx}$$$$r^2+8r+12=0$$$$(r+2)(r+6)=0$$$$\Rightarrow z=A_1{e}^{-2x}+B_1{e}^{-6x}=y^{\prime }$$$$\Rightarrow y=\frac{A_1}{-2}{e}^{-2x}+\frac{B_1}{-6}{e}^{-6x}+C$$$$\Rightarrow y={Ae}^{-2x}+{Be}^{-6x}+C$$

Commented by Mastermind last updated on 23/Dec/22

$$\mathrm{There}\mathrm{was}\mathrm{an}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{question},$$$${\mathrm{i}}^{\prime }\mathrm{m}\mathrm{sorry}$$$$\mathrm{i}\mathrm{have}\mathrm{corrected}\mathrm{it}\mathrm{tho}.$$

Commented by aleks041103 last updated on 24/Dec/22

$$I{}^{\prime }vecorrectedmyansweralso$$

Commented by Mastermind last updated on 27/Dec/22

$$\mathrm{How}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{your}\mathrm{last}\mathrm{answer}?$$

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