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d-2-y-dx-2-4y-tan-2x-




Question Number 9396 by Shiva last updated on 04/Dec/16
(d^2 y/dx^2 )+4y=tan 2x
$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{4y}=\mathrm{tan}\:\mathrm{2x} \\ $$
Commented by 123456 last updated on 04/Dec/16
y=y_n +y_f   y_n :  (d^2 y_n /dx^2 )+4y_n =0  y_n =Ae^(λt)   λ^2 +4=0⇒λ=±2ι  y_n =C_1 sin (2t)+C_2 cos (2t)
$${y}={y}_{{n}} +{y}_{{f}} \\ $$$${y}_{{n}} : \\ $$$$\frac{{d}^{\mathrm{2}} {y}_{{n}} }{{dx}^{\mathrm{2}} }+\mathrm{4}{y}_{{n}} =\mathrm{0} \\ $$$${y}_{{n}} ={Ae}^{\lambda{t}} \\ $$$$\lambda^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\Rightarrow\lambda=\pm\mathrm{2}\iota \\ $$$${y}_{{n}} ={C}_{\mathrm{1}} \mathrm{sin}\:\left(\mathrm{2}{t}\right)+{C}_{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{2}{t}\right) \\ $$
Answered by Raja Naik last updated on 09/Dec/16
(D^2 +4)y=tan2x→(1)  f(D)=D^2 +4      Q(x)=tan2x  y_c =A.E is f(m)=0  m^2 +4=0  m^2 +2^2 =0  m^2 =−2^2   m=±2i  y_c =c_(1 ) cos2x+c_2 sin2x→(2)  to find y_p =  (1/(f(D)))Q  (1/(D^2 +2^2 ))tan2x  (1/(2ai))((1/(D−2i))−(1/(D+2i)))tan2x→(3)  consider,  (1/(D−2i))tan2x  e^(i2x) ∫e^(−i2x) .tan2x dx  e^(i2x) ∫(cos2x−isin2x)((sin2x)/(cos2x))dx  e^(i2x) ∫[sin2x−i(((1−cos^2 2x))/(cos2x)) ]dx  e^(i2x) [∫sin2xdx−isec2xdx+i∫cos2xdx  e^(i2x) [−((cos2x)/2) −(i/2)log∣tan(Π/4)+((2x)/2)∣+(i/2)sin2x]  −(e^(i2x) /2)[(cos2x−isin2x)+ilog((Π/4)+x)]  −(e^(i2x) /2)[e^(−i2x) +ilog tan((Π/4)+x)]  thus (1/(D−2i))tan2x=−(1/2)−(i/2)e^(i2x) logtan((Π/4)+x)→(4)  replace i by −i we get  (1/(D+2i ))tan2x=−(1/2)+(i/2)e^(−i2x) logtan((Π/4)+x)→(5)  substitide (4),(5) in (3) we get  y_p =(1/i)[−(i/2)−(i/2)e^(i2x) log tan((Π/4)+x)+(1/2)−(i/2)e^(−i2x) log tan((Π/4)+x)]  y_p =(1/i)[−(((e^(i2x) +e^(−i2x) ))/2)log tan ((Π/4)+x)]  y_p =−cos2x log tan((Π/4)+x)→(6)    the General solution is  y=y_c +y_p   y=(2)+(6)  y=c_1 cos2x+c_2 sin2x−cos2x log tan((Π/4)+x)           −Raja Naik.
$$\left(\mathrm{D}^{\mathrm{2}} +\mathrm{4}\right)\mathrm{y}=\mathrm{tan2x}\rightarrow\left(\mathrm{1}\right) \\ $$$$\mathrm{f}\left(\mathrm{D}\right)=\mathrm{D}^{\mathrm{2}} +\mathrm{4}\:\:\:\:\:\:\mathrm{Q}\left(\mathrm{x}\right)=\mathrm{tan2x} \\ $$$$\mathrm{y}_{\mathrm{c}} =\mathrm{A}.\mathrm{E}\:\mathrm{is}\:\mathrm{f}\left(\mathrm{m}\right)=\mathrm{0} \\ $$$$\mathrm{m}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{m}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{m}^{\mathrm{2}} =−\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{m}=\pm\mathrm{2i} \\ $$$$\mathrm{y}_{\mathrm{c}} =\mathrm{c}_{\mathrm{1}\:} \mathrm{cos2x}+\mathrm{c}_{\mathrm{2}} \mathrm{sin2x}\rightarrow\left(\mathrm{2}\right) \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{y}_{\mathrm{p}} = \\ $$$$\frac{\mathrm{1}}{\mathrm{f}\left(\mathrm{D}\right)}\mathrm{Q} \\ $$$$\frac{\mathrm{1}}{\mathrm{D}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\mathrm{tan2x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2ai}}\left(\frac{\mathrm{1}}{\mathrm{D}−\mathrm{2i}}−\frac{\mathrm{1}}{\mathrm{D}+\mathrm{2i}}\right)\mathrm{tan2x}\rightarrow\left(\mathrm{3}\right) \\ $$$$\mathrm{consider}, \\ $$$$\frac{\mathrm{1}}{\mathrm{D}−\mathrm{2i}}\mathrm{tan2x} \\ $$$$\mathrm{e}^{\mathrm{i2x}} \int\mathrm{e}^{−\mathrm{i2x}} .\mathrm{tan2x}\:\mathrm{dx} \\ $$$$\mathrm{e}^{\mathrm{i2x}} \int\left(\mathrm{cos2x}−\mathrm{isin2x}\right)\frac{\mathrm{sin2x}}{\mathrm{cos2x}}\mathrm{dx} \\ $$$$\mathrm{e}^{\mathrm{i2x}} \int\left[\mathrm{sin2x}−\mathrm{i}\frac{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{2x}\right)}{\mathrm{cos2x}}\:\right]\mathrm{dx} \\ $$$$\mathrm{e}^{\mathrm{i2x}} \left[\int\mathrm{sin2xdx}−\mathrm{isec2xdx}+\mathrm{i}\int\mathrm{cos2xdx}\right. \\ $$$$\mathrm{e}^{\mathrm{i2x}} \left[−\frac{\mathrm{cos2x}}{\mathrm{2}}\:−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{log}\mid\mathrm{tan}\frac{\Pi}{\mathrm{4}}+\frac{\mathrm{2x}}{\mathrm{2}}\mid+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{sin2x}\right] \\ $$$$−\frac{\mathrm{e}^{\mathrm{i2x}} }{\mathrm{2}}\left[\left(\mathrm{cos2x}−\mathrm{isin2x}\right)+\mathrm{ilog}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\right] \\ $$$$−\frac{\mathrm{e}^{\mathrm{i2x}} }{\mathrm{2}}\left[\mathrm{e}^{−\mathrm{i2x}} +\mathrm{ilog}\:\mathrm{tan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\right] \\ $$$$\mathrm{thus}\:\frac{\mathrm{1}}{\mathrm{D}−\mathrm{2i}}\mathrm{tan2x}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{e}^{\mathrm{i2x}} \mathrm{logtan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\rightarrow\left(\mathrm{4}\right) \\ $$$$\mathrm{replace}\:\mathrm{i}\:\mathrm{by}\:−\mathrm{i}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{D}+\mathrm{2i}\:}\mathrm{tan2x}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i2x}} \mathrm{logtan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\rightarrow\left(\mathrm{5}\right) \\ $$$$\mathrm{substitide}\:\left(\mathrm{4}\right),\left(\mathrm{5}\right)\:\mathrm{in}\:\left(\mathrm{3}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{y}_{\mathrm{p}} =\frac{\mathrm{1}}{\mathrm{i}}\left[−\frac{\mathrm{i}}{\mathrm{2}}−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{e}^{\mathrm{i2x}} \mathrm{log}\:\mathrm{tan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i2x}} \mathrm{log}\:\mathrm{tan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\right] \\ $$$$\mathrm{y}_{\mathrm{p}} =\frac{\mathrm{1}}{\mathrm{i}}\left[−\frac{\left(\mathrm{e}^{\mathrm{i2x}} +\mathrm{e}^{−\mathrm{i2x}} \right)}{\mathrm{2}}\mathrm{log}\:\mathrm{tan}\:\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\right] \\ $$$$\mathrm{y}_{\mathrm{p}} =−\mathrm{cos2x}\:\mathrm{log}\:\mathrm{tan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right)\rightarrow\left(\mathrm{6}\right) \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{General}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{y}=\mathrm{y}_{\mathrm{c}} +\mathrm{y}_{\mathrm{p}} \\ $$$$\mathrm{y}=\left(\mathrm{2}\right)+\left(\mathrm{6}\right) \\ $$$$\mathrm{y}=\mathrm{c}_{\mathrm{1}} \mathrm{cos2x}+\mathrm{c}_{\mathrm{2}} \mathrm{sin2x}−\mathrm{cos2x}\:\mathrm{log}\:\mathrm{tan}\left(\frac{\Pi}{\mathrm{4}}+\mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:−\mathrm{Raja}\:\mathrm{Naik}. \\ $$

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