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Question Number 67759 by ugwu Kingsley last updated on 31/Aug/19
using variation of parameters method (x+2)^2 y′′−(x+2)y′=2x+4 x^2 y′′+2xy′−2y=x^2 lnx+3x
usingvariationofparametersmethod(x+2)2y″−(x+2)y′=2x+4x2y″+2xy′−2y=x2lnx+3x
Commented by mathmax by abdo last updated on 31/Aug/19
1) changement y^′ =z give (x+2)^2 z′−(x+2)z =2x+4 (he)→(x+2)^2 z^′ −(x+2)z =0 ⇒(x+2)^2 z^′ =(x+2)z ⇒ (z^′ /z) =((x+2)/((x+2)^2 )) ⇒(z^′ /z) =(1/(x+2)) ⇒ln∣z∣=ln∣x+2∣+c ⇒z =K∣x+2∣let find the solution on]−2,+∞[let use mvc method ⇒z^′ =K^′ (x+2)+K (e) ⇒(x+2)^2 K^′ (x+2) +K(x+2)^2 −(x+2)K(x+2) =2x+4 ⇒ (x+2)^3 K^′ =2(x+2) ⇒K^′ =(2/((x+2)^2 )) ⇒K(x)=−(2/(x+2)) +c ⇒ z(x) =(−(2/(x+2))+c)(x+2) =−2 +c y^′ =z ⇒y(x) =∫z(x) dx +λ =∫(−2+c)dx +λ =(−2+c)x +λ
1)changementy′=zgive(x+2)2z′−(x+2)z=2x+4(he)→(x+2)2z′−(x+2)z=0⇒(x+2)2z′=(x+2)z⇒z′z=x+2(x+2)2⇒z′z=1x+2⇒ln∣z∣=ln∣x+2∣+c⇒z=K∣x+2∣letfindthesolutionon]−2,+∞[letusemvcmethod⇒z′=K′(x+2)+K(e)⇒(x+2)2K′(x+2)+K(x+2)2−(x+2)K(x+2)=2x+4⇒(x+2)3K′=2(x+2)⇒K′=2(x+2)2⇒K(x)=−2x+2+c⇒z(x)=(−2x+2+c)(x+2)=−2+cy′=z⇒y(x)=∫z(x)dx+λ=∫(−2+c)dx+λ=(−2+c)x+λ

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