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Question Number 202651 by hardmath last updated on 31/Dec/23
Commented by Frix last updated on 31/Dec/23
y=((3cos x +sin x)/(10))+c_1 e^(2x) +c_2 e^x
$${y}=\frac{\mathrm{3cos}\:{x}\:+\mathrm{sin}\:{x}}{\mathrm{10}}+{c}_{\mathrm{1}} \mathrm{e}^{\mathrm{2}{x}} +{c}_{\mathrm{2}} \mathrm{e}^{{x}} \\ $$
Answered by Mathspace last updated on 31/Dec/23
h→r^2 −3r+2=0 Δ=9−8=1 ⇒r_1 =((3+1)/2)=2 r_2 =((3−1)/2)=1 ⇒y_h =ae^x +be^(2x) =au_1 +bu_2 w(u_1 ,u_2 )= determinant (((u_1 u_2 )),((u_1 ^′ u_2 ^′ ))) = determinant (((e^x e^(2x) )),((e^x 2e^(2x) )))=2e^(3x) −e^(3x) =e^(3x) w_1 = determinant (((0 e^(2x) )),((sinx 2e^(2x) )))=−e^(2x) sinx w_2 = determinant (((e^x 0)),((e^x sinx)))=e^x sinx v_1 =∫ (w_1 /w)dx=−∫((e^(2x) sinx)/e^(3x) )dx =−∫e^(−x) sinxdx =−Im(∫ e^(−x+ix) dx) we have ∫ e^((−1+i)x) dx =(1/(−1+i))e^((−1+i)x) =−(1/(1−i))e^((−1+i)x) =−((1+i)/2)e^(−x) (cosx+isinx) =−(e^(−x) /2)(cosx+isinx−icosx−sinx) ⇒v_1 =(e^(−x) /2)(sinx−cosx) v_2 =∫(w_2 /w)dx=∫((e^x sinx)/e^(3x) )dx ∫ e^(−2x) sinx dx=Im(∫e^((−2+i)x) dx) ∫ e^((−2+i)x) dx=(1/(−2+i))e^((−2+i)x) =−(1/(2−i))e^((−2+i)x) =−((2+i)/5)e^(−2x) (cosx+isinx) =−(e^(−2x) /5)(2cosx+2isinx+icosx−sinx) ⇒v_2 =−(e^(−2x) /5)(2sinx+cosx) particular solution is y_p =u_1 v_1 +u_2 v_2 =e^x .(e^(−x) /2)(sinx−cosx) −e^(2x) .(e^(−2x) /5)(2sinx+cosx) =(1/2)sinx−(1/2)cosx−(2/5)sinx −(1/5)cosx=(1/(10))sinx−(7/(10))cosx so y_g =y_p +y_h =(1/(10))sinx−(7/(10))cosx +ae^x +be^(2x)
$${h}\rightarrow{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\Delta=\mathrm{9}−\mathrm{8}=\mathrm{1}\:\Rightarrow{r}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{1}}{\mathrm{2}}=\mathrm{2} \\ $$$${r}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\Rightarrow{y}_{{h}} ={ae}^{{x}} +{be}^{\mathrm{2}{x}} \\ $$$$={au}_{\mathrm{1}} +{bu}_{\mathrm{2}} \\ $$$${w}\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \right)=\begin{vmatrix}{{u}_{\mathrm{1}} \:\:\:\:\:\:{u}_{\mathrm{2}} }\\{{u}_{\mathrm{1}} ^{'} \:\:\:\:\:\:\:{u}_{\mathrm{2}} ^{'} }\end{vmatrix} \\ $$$$=\begin{vmatrix}{{e}^{{x}} \:\:\:\:\:\:\:\:\:{e}^{\mathrm{2}{x}} }\\{{e}^{{x}} \:\:\:\:\:\:\:\:\:\mathrm{2}{e}^{\mathrm{2}{x}} }\end{vmatrix}=\mathrm{2}{e}^{\mathrm{3}{x}} −{e}^{\mathrm{3}{x}} ={e}^{\mathrm{3}{x}} \\ $$$${w}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:{e}^{\mathrm{2}{x}} }\\{{sinx}\:\:\:\mathrm{2}{e}^{\mathrm{2}{x}} }\end{vmatrix}=−{e}^{\mathrm{2}{x}} {sinx} \\ $$$${w}_{\mathrm{2}} =\begin{vmatrix}{{e}^{{x}} \:\:\:\:\:\:\:\mathrm{0}}\\{{e}^{{x}} \:\:\:\:\:\:{sinx}}\end{vmatrix}={e}^{{x}} {sinx} \\ $$$${v}_{\mathrm{1}} =\int\:\frac{{w}_{\mathrm{1}} }{{w}}{dx}=−\int\frac{{e}^{\mathrm{2}{x}} {sinx}}{{e}^{\mathrm{3}{x}} }{dx} \\ $$$$=−\int{e}^{−{x}} {sinxdx} \\ $$$$=−{Im}\left(\int\:{e}^{−{x}+{ix}} {dx}\right) \\ $$$${we}\:{have}\:\int\:{e}^{\left(−\mathrm{1}+{i}\right){x}} {dx} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{1}+{i}}{e}^{\left(−\mathrm{1}+{i}\right){x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{1}−{i}}{e}^{\left(−\mathrm{1}+{i}\right){x}} =−\frac{\mathrm{1}+{i}}{\mathrm{2}}{e}^{−{x}} \left({cosx}+{isinx}\right) \\ $$$$=−\frac{{e}^{−{x}} }{\mathrm{2}}\left({cosx}+{isinx}−{icosx}−{sinx}\right) \\ $$$$\Rightarrow{v}_{\mathrm{1}} =\frac{{e}^{−{x}} }{\mathrm{2}}\left({sinx}−{cosx}\right) \\ $$$${v}_{\mathrm{2}} =\int\frac{{w}_{\mathrm{2}} }{{w}}{dx}=\int\frac{{e}^{{x}} {sinx}}{{e}^{\mathrm{3}{x}} }{dx} \\ $$$$\int\:{e}^{−\mathrm{2}{x}} {sinx}\:{dx}={Im}\left(\int{e}^{\left(−\mathrm{2}+{i}\right){x}} {dx}\right) \\ $$$$\int\:{e}^{\left(−\mathrm{2}+{i}\right){x}} {dx}=\frac{\mathrm{1}}{−\mathrm{2}+{i}}{e}^{\left(−\mathrm{2}+{i}\right){x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}−{i}}{e}^{\left(−\mathrm{2}+{i}\right){x}} =−\frac{\mathrm{2}+{i}}{\mathrm{5}}{e}^{−\mathrm{2}{x}} \left({cosx}+{isinx}\right) \\ $$$$=−\frac{{e}^{−\mathrm{2}{x}} }{\mathrm{5}}\left(\mathrm{2}{cosx}+\mathrm{2}{isinx}+{icosx}−{sinx}\right) \\ $$$$\Rightarrow{v}_{\mathrm{2}} =−\frac{{e}^{−\mathrm{2}{x}} }{\mathrm{5}}\left(\mathrm{2}{sinx}+{cosx}\right) \\ $$$${particular}\:{solution}\:{is} \\ $$$${y}_{{p}} ={u}_{\mathrm{1}} {v}_{\mathrm{1}} +{u}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$$={e}^{{x}} .\frac{{e}^{−{x}} }{\mathrm{2}}\left({sinx}−{cosx}\right) \\ $$$$−{e}^{\mathrm{2}{x}} .\frac{{e}^{−\mathrm{2}{x}} }{\mathrm{5}}\left(\mathrm{2}{sinx}+{cosx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sinx}−\frac{\mathrm{1}}{\mathrm{2}}{cosx}−\frac{\mathrm{2}}{\mathrm{5}}{sinx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{5}}{cosx}=\frac{\mathrm{1}}{\mathrm{10}}{sinx}−\frac{\mathrm{7}}{\mathrm{10}}{cosx} \\ $$$${so}\:{y}_{{g}} ={y}_{{p}} +{y}_{{h}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}{sinx}−\frac{\mathrm{7}}{\mathrm{10}}{cosx}\:+{ae}^{{x}} +{be}^{\mathrm{2}{x}} \\ $$
Commented by Frix last updated on 03/Jan/24
The method is ok but the result is wrong.
$$\mathrm{The}\:\mathrm{method}\:\mathrm{is}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{wrong}. \\ $$

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