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Question Number 38802 by tawa tawa last updated on 30/Jun/18

solve:   y′′(1 + 4x^2 ) − 8y = 0

$$\mathrm{solve}:\:\:\:\mathrm{y}''\left(\mathrm{1}\:+\:\mathrm{4x}^{\mathrm{2}} \right)\:−\:\mathrm{8y}\:=\:\mathrm{0} \\ $$

Answered by MrW3 last updated on 30/Jun/18

y=ax^2 +bx+c  y′=2ax+b  y′′=2a  2a(1+4x^2 )−8(ax^2 +bx+c)=0  8ax^2 +2a−8ax^2 −8bx−8c=0  −8bx+2a−8c=0  ⇒−8b=0⇒b=0  ⇒2a−8c=0⇒a=4c  ⇒y=c(4x^2 +1)

$${y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${y}'=\mathrm{2}{ax}+{b} \\ $$$${y}''=\mathrm{2}{a} \\ $$$$\mathrm{2}{a}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)−\mathrm{8}\left({ax}^{\mathrm{2}} +{bx}+{c}\right)=\mathrm{0} \\ $$$$\mathrm{8}{ax}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{8}{ax}^{\mathrm{2}} −\mathrm{8}{bx}−\mathrm{8}{c}=\mathrm{0} \\ $$$$−\mathrm{8}{bx}+\mathrm{2}{a}−\mathrm{8}{c}=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{8}{b}=\mathrm{0}\Rightarrow{b}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}−\mathrm{8}{c}=\mathrm{0}\Rightarrow{a}=\mathrm{4}{c} \\ $$$$\Rightarrow{y}={c}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$

Commented by tawa tawa last updated on 30/Jun/18

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18

y=x^α +h by trial method  y=x^2 +(1/4) is a solution...  (dy/dx)=2x  (d^2 y/dx^2 )=2  2(1+4x^2 )−8(x^2 +(1/4))=0  SO y=x^2 +(1/4)  is the solution  pls give me time to solve by standard method

$${y}={x}^{\alpha} +{h}\:{by}\:{trial}\:{method} \\ $$$${y}={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\:{is}\:{a}\:{solution}... \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{x}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)−\mathrm{8}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{S}}{O}\:{y}={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\:\:{is}\:{the}\:{solution} \\ $$$${pls}\:{give}\:{me}\:{time}\:{to}\:{solve}\:{by}\:{standard}\:{method} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by tawa tawa last updated on 30/Jun/18

i will be waiting sir. God bless you

$$\mathrm{i}\:\mathrm{will}\:\mathrm{be}\:\mathrm{waiting}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

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