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Question Number 26073 by gopikrishnan005@gmail.com last updated on 19/Dec/17

  solve the differential equation(D^2 +2D+1)y=x^2 +2x+1

$$ \\ $$$${solve}\:{the}\:{differential}\:{equation}\left({D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{1}\right){y}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$

Commented by gopikrishnan005@gmail.com last updated on 20/Dec/17

pls explain

$${pls}\:{explain} \\ $$

Commented by abdo imad last updated on 21/Dec/17

⇔  (d^2 y/dx^2 )  +2(dy/dx)  +y =x^2 +2x+1  the solution of this ED is  y_g   = y_h  +y_p   .. y_h   is the homogen s^t   and y_p   particular s^t   EH−−> (d^2 y/dx^2 ) + 2(dy/dx)  +1 =0 have the caracteristic equation  x^2  +2x +1=0⇔  (x+1)^2 =0⇔ x=−1⇒  y_h   =(αx+β)e^(−x)  (−1 is double root)  for y_p   we put y=ax^4  +bx^3  +cx^2  +dx +e⇒  (dy/dx)= 4ax^3 + 3bx^2  +2cx +d  and  (d^2 y/dx^2 )  = 12ax^2 +6bx +2c  we find after calculus  a=0..b=0...c=1..d=−2...e=3  and y_g   =(αx+β)e^(−x)   +x^2  −2x +3  .

$$\Leftrightarrow\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:\:+\mathrm{2}\frac{{dy}}{{dx}}\:\:+{y}\:={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\:\:{the}\:{solution}\:{of}\:{this}\:{ED}\:{is} \\ $$$${y}_{{g}} \:\:=\:{y}_{{h}} \:+{y}_{{p}} \:\:..\:{y}_{{h}} \:\:{is}\:{the}\:{homogen}\:{s}^{{t}} \:\:{and}\:{y}_{{p}} \:\:{particular}\:{s}^{{t}} \\ $$$${EH}−−>\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\frac{{dy}}{{dx}}\:\:+\mathrm{1}\:=\mathrm{0}\:{have}\:{the}\:{caracteristic}\:{equation} \\ $$$${x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}=\mathrm{0}\Leftrightarrow\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Leftrightarrow\:{x}=−\mathrm{1}\Rightarrow\:\:{y}_{{h}} \:\:=\left(\alpha{x}+\beta\right){e}^{−{x}} \:\left(−\mathrm{1}\:{is}\:{double}\:{root}\right) \\ $$$${for}\:{y}_{{p}} \:\:{we}\:{put}\:{y}={ax}^{\mathrm{4}} \:+{bx}^{\mathrm{3}} \:+{cx}^{\mathrm{2}} \:+{dx}\:+{e}\Rightarrow \\ $$$$\frac{{dy}}{{dx}}=\:\mathrm{4}{ax}^{\mathrm{3}} +\:\mathrm{3}{bx}^{\mathrm{2}} \:+\mathrm{2}{cx}\:+{d}\:\:{and}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:\:=\:\mathrm{12}{ax}^{\mathrm{2}} +\mathrm{6}{bx}\:+\mathrm{2}{c} \\ $$$${we}\:{find}\:{after}\:{calculus}\:\:{a}=\mathrm{0}..{b}=\mathrm{0}...{c}=\mathrm{1}..{d}=−\mathrm{2}...{e}=\mathrm{3} \\ $$$${and}\:{y}_{{g}} \:\:=\left(\alpha{x}+\beta\right){e}^{−{x}} \:\:+{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:+\mathrm{3}\:\:. \\ $$

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