Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 127589 by bramlexs22 last updated on 31/Dec/20

Answered by liberty last updated on 31/Dec/20

(•) let y′=z ⇒ xz′ + z = 3x^2 −x       (d/dx) (xz) = 3x^2 −x      xz = x^3 −(1/2)x^2  +C_1       z = x^2 −(1/2)x + (C_1 /x)      (dy/dx) = x^2 −(1/2)x+ (C_1 /x)      y = (1/3)x^3 −(1/4)x^2 + C_1 ln ∣x∣ + C_2   apply the initial condition y(0)=1 and y′(0)=1    start with the first derivative    0.1 = 0^3 −(1/2).0^2  + C_1  ⇒C_1 =0   and 1 = (0^3 /3)−(0^4 /4)+C_2  ⇒C_2 =1  ∴ y = (x^3 /3)−(x^2 /4)+1

$$\left(\bullet\right)\:\mathrm{let}\:\mathrm{y}'=\mathrm{z}\:\Rightarrow\:\mathrm{xz}'\:+\:\mathrm{z}\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{xz}\right)\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\mathrm{xz}\:=\:\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:+\mathrm{C}_{\mathrm{1}} \\ $$$$\:\:\:\:\mathrm{z}\:=\:\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:+\:\frac{\mathrm{C}_{\mathrm{1}} }{\mathrm{x}} \\ $$$$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\:\frac{\mathrm{C}_{\mathrm{1}} }{\mathrm{x}} \\ $$$$\:\:\:\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} +\:\mathrm{C}_{\mathrm{1}} \mathrm{ln}\:\mid\mathrm{x}\mid\:+\:\mathrm{C}_{\mathrm{2}} \\ $$$$\mathrm{apply}\:\mathrm{the}\:\mathrm{initial}\:\mathrm{condition}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\:\:\mathrm{start}\:\mathrm{with}\:\mathrm{the}\:\mathrm{first}\:\mathrm{derivative}\: \\ $$$$\:\mathrm{0}.\mathrm{1}\:=\:\mathrm{0}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{0}^{\mathrm{2}} \:+\:\mathrm{C}_{\mathrm{1}} \:\Rightarrow\mathrm{C}_{\mathrm{1}} =\mathrm{0}\: \\ $$$$\mathrm{and}\:\mathrm{1}\:=\:\frac{\mathrm{0}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{0}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{C}_{\mathrm{2}} \:\Rightarrow\mathrm{C}_{\mathrm{2}} =\mathrm{1} \\ $$$$\therefore\:\mathrm{y}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com