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Question Number 271 by 123456 last updated on 25/Jan/15

x^3 (d^2 y/dx^2 )+x^2 (dy/dx)+xy=2−x y(1)=0 y′(1)=−1

$$\:{x}^{\mathrm{3}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}^{\mathrm{2}} \frac{{dy}}{{dx}}+{xy}=\mathrm{2}−{x} \\ $$$${y}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${y}'\left(\mathrm{1}\right)=−\mathrm{1} \\ $$

Answered by prakash jain last updated on 18/Dec/14

Homegenous Solution y=x^r y′=rx^(r−1) y′′=r(r−1)x^(r−2) r(r−1)x^(r+1) +rx^(r+1) +x^(r+1) =0 r^2 −r+r+1=0 r=i, −i y_h =C_1 cos (ln ∣x∣)+C_2 sin (ln ∣x∣) Particular Solution y=(A/x)+B y′=−(A/x^2 ) y′′=((2A)/x^3 ) x^3 y′′+x^2 y′+xy=2−x 2A−A+A+Bx=2−x A=1, B=−1 Solution y=y_h +y_p y=C_1 cos (ln ∣x∣)+C_2 sin (ln ∣x∣)+(1/x)−1 Solution under the given condition Given y(1)=1 1=C_1 y′(1)=−1 y′=−(1/x)sin (ln ∣x∣)+(C_2 /x)cos (ln ∣x∣)−(1/x^2 ) −1=C_2 −1 C_2 =0 y=cos (ln ∣x∣)+(1/x)−1

$$\mathrm{Homegenous}\:\mathrm{Solution} \\ $$$${y}={x}^{{r}} \\ $$$${y}'={rx}^{{r}−\mathrm{1}} \:\:{y}''={r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{2}} \\ $$$${r}\left({r}−\mathrm{1}\right){x}^{{r}+\mathrm{1}} +{rx}^{{r}+\mathrm{1}} +{x}^{{r}+\mathrm{1}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}+{r}+\mathrm{1}=\mathrm{0} \\ $$$${r}={i},\:−{i} \\ $$$${y}_{{h}} ={C}_{\mathrm{1}} \mathrm{cos}\:\left(\mathrm{ln}\:\mid{x}\mid\right)+{C}_{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:\mid{x}\mid\right) \\ $$$$\mathrm{Particular}\:\mathrm{Solution} \\ $$$${y}=\frac{{A}}{{x}}+{B} \\ $$$${y}'=−\frac{{A}}{{x}^{\mathrm{2}} } \\ $$$${y}''=\frac{\mathrm{2}{A}}{{x}^{\mathrm{3}} } \\ $$$${x}^{\mathrm{3}} {y}''+{x}^{\mathrm{2}} {y}'+{xy}=\mathrm{2}−{x} \\ $$$$\mathrm{2}{A}−{A}+{A}+{Bx}=\mathrm{2}−{x} \\ $$$${A}=\mathrm{1},\:{B}=−\mathrm{1} \\ $$$$\mathrm{Solution} \\ $$$${y}={y}_{{h}} +{y}_{{p}} \\ $$$${y}={C}_{\mathrm{1}} \mathrm{cos}\:\left(\mathrm{ln}\:\mid{x}\mid\right)+{C}_{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:\mid{x}\mid\right)+\frac{\mathrm{1}}{{x}}−\mathrm{1} \\ $$$$\mathrm{Solution}\:\mathrm{under}\:\mathrm{the}\:\mathrm{given}\:\mathrm{condition} \\ $$$$\mathrm{Given}\:{y}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{1}={C}_{\mathrm{1}} \\ $$$${y}'\left(\mathrm{1}\right)=−\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{1}}{{x}}\mathrm{sin}\:\left(\mathrm{ln}\:\mid{x}\mid\right)+\frac{{C}_{\mathrm{2}} }{{x}}\mathrm{cos}\:\left(\mathrm{ln}\:\mid{x}\mid\right)−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$−\mathrm{1}={C}_{\mathrm{2}} −\mathrm{1} \\ $$$${C}_{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\mathrm{cos}\:\left(\mathrm{ln}\:\mid{x}\mid\right)+\frac{\mathrm{1}}{{x}}−\mathrm{1} \\ $$

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