Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 82247 by jagoll last updated on 19/Feb/20

$$(D^4+2D^2+1)y=x^2\cos x$$

Commented by john santu last updated on 20/Feb/20

$$characteristicequation$$$${\lambda }^4+2{\lambda }^2+1=0$$$${({\lambda }^2+1)}^2=0\Rightarrow {({\lambda }^2-i)}^2{({\lambda }^2+i)}^2=0$$$${\lambda }^2=i\Rightarrow {\lambda }_{1,2}=\pm \sqrt{i}$$$${\lambda }^2=-i\Rightarrow {\lambda }_{3,4}=\pm \sqrt{-i}$$$$y_h=Ae{}^{\sqrt{i}}+{Be}^{-\sqrt{i}}+{Ce}^{-\sqrt{-i}}+{De}^{\sqrt{-i}}$$

Answered by TANMAY PANACEA last updated on 20/Feb/20

$$y=\frac{x^{2}cosx}{{(D^2+1)}^2}$$$$y_{real}+y_{imaginary}=\frac{x^2{e}^{ix}}{{(D^2+1)}^2}$$$$=\frac{e^{ix}}{{[{(D+i)}^2+1]}^2}\times x^2$$$$=e^{ix}\times \frac{1}{{(D^2+2iD)}^2}\times x^2$$$$=e^{ix}\times \frac{1}{D^2}\times \frac{1}{{(D+2i)}^2}\times x^2$$$$=e^{ix}\times \frac{1}{-4{(1+\frac{D}{2i})}^2}\times \frac{x^4}{12}[\frac{1}{D^2}\times x^2=\frac{1}{D}\times \frac{x^3}{3}=\frac{x^4}{12}]$$$$nowusing$$$${(1-a)}^{-2}=1+2a+3a^2+4a^3+5a^4+6a^5+...+(r+1)a^r$$$$puta=-\frac{D}{2i}$$$$=\frac{(cosx+isinx)}{-48}\times (1-2.\frac{D}{2i}+3\times \frac{D^2}{-4}+4\times \frac{D^3}{8i}+5\times \frac{D^4}{16}otherstermsignored)x^4$$$$=\frac{(cosx+isinx)}{-48}(x^4-\frac{4x^3}{i}-\frac{3}{4}\times 12x^2+\frac{1}{2i}\times 24x+\frac{5}{16}\times 24)$$$$=\frac{cosx+isinx}{-48}(x^4-9x^2+\frac{15}{2}+i\times 4x^3-i\times 12x)$$$$requiredanswerisrealpart$$$$=\frac{cosx}{-48}(x^4-9x^2+\frac{15}{2})+\frac{sinx}{-48}\times (-1)(4x^3-12x)$$$$plscheckmistakdifany$$

Commented by jagoll last updated on 20/Feb/20

$$thanksir$$

Commented by TANMAY PANACEA last updated on 20/Feb/20

$$mostwelcome$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com