Second order linear equations #1 - Questions and Answers

Question 1 Use the method of variation of parameters to find a particular solution of the given differential equation. 𝑦 ′ + 𝑦 = csc ⁡ 2 𝑥 y ′ +y=csc 2 x Answer Take a look at the differential equation: 𝑦 ′ + 𝑦 = csc ⁡ 2 𝑥 y ′ +y=csc 2 x Finding a certain solution is the goal, 𝑦 𝑝 y p ​ , of the given equation. The homogeneous equation connected to the previous equation is: 𝑦 ′ + 𝑦 = 0 y ′ +y=0 Equation that is auxiliary to the homogeneous differential equation is: 𝑟 2 + 1 = 0 r 2 +1=0 𝑟 = ± 𝑖 r=±i 𝑟 = ± 1 𝑖 r=±1i Consequently, the auxiliary equation’s roots are: 𝑟 1 = 𝑖 ,   𝑟 2 = − 𝑖 r 1 ​ =i,r 2 ​ =−i Consequently, the complementary role is: 𝑦 𝑐 = 𝑐 1 cos ⁡ 𝑥 + 𝑐 2 sin ⁡ 𝑥 y c ​ =c 1 ​ cosx+c 2 ​ sinx We have: 𝑦 1 ( 𝑥 ) = cos ⁡ 𝑥 ,   𝑦 2 ( 𝑥 ) = sin ⁡ 𝑥 y 1 ​ (x)=cosx,y 2 ​ (x)=sinx 𝑦 1 ′ ( 𝑥 ) = − sin ⁡ 𝑥 ,   𝑦 2 ′ ( 𝑥 ) = cos ⁡ 𝑥 y 1 ′ ​ (x)=−sinx,y 2 ′ ​ (x)=cosx Hence: 𝑊 = Wronskian = ∣ cos ⁡ 𝑥 sin ⁡ 𝑥 − sin ⁡ 𝑥 cos ⁡ 𝑥 ∣ W=Wronskian= ​ cosx −sinx ​ sinx cosx ​ ​ 𝑊 = cos ⁡ 2 𝑥 + sin ⁡ 2 𝑥 = 1 W=cos 2 x+sin 2 x=1 Here, 𝑓 ( 𝑥 ) = csc ⁡ 2 𝑥 f(x)=csc 2 x. The specific response is provided by: 𝑦 𝑝 ( 𝑥 ) = ( − ∫ 𝑦 2 ( 𝑥 ) 𝑓 ( 𝑥 ) 𝑊 ( 𝑥 )   𝑑 𝑥 ⋅ 𝑦 1 ( 𝑥 ) ) + ( ∫ 𝑦 1 ( 𝑥 ) 𝑓 ( 𝑥 ) 𝑊 ( 𝑥 )   𝑑 𝑥 ⋅ 𝑦 2 ( 𝑥 ) ) y p ​ (x)=(−∫ W(x) y 2 ​ (x)f(x) ​ dx⋅y 1 ​ (x))+(∫ W(x) y 1 ​ (x)f(x) ​ dx⋅y 2 ​ (x)) 𝑦 𝑝 ( 𝑥 ) = ( − ∫ sin ⁡ 𝑥 csc ⁡ 2 𝑥   𝑑 𝑥 ⋅ cos ⁡ 𝑥 ) + ( ∫ cos ⁡ 𝑥 csc ⁡ 2 𝑥   𝑑 𝑥 ⋅ sin ⁡ 𝑥 ) y p ​ (x)=(−∫sinxcsc 2 xdx⋅cosx)+(∫cosxcsc 2 xdx⋅sinx) Question 2 If 𝑥 𝑦 + 6 𝑒 𝑦 = 6 𝑒 xy+6e y =6e, find the value of 𝑦 ′ y ′ at the point where 𝑥 = 0 x=0. Answer Here the differential equation is given by: 𝑥 𝑦 + 6 𝑒 𝑦 = 6 𝑒 with 𝑦 ( 0 ) = 1 xy+6e y =6ewithy(0)=1 Taking the derivative of both sides with respect to 𝑥 x, we get: ( 𝑑 / 𝑑 𝑥 ) ( 𝑥 𝑦 + 6 𝑒 𝑦 ) = ( 𝑑 / 𝑑 𝑥 ) ( 6 𝑒 ) (d/dx)(xy+6e y )=(d/dx)(6e) 𝑦 + 𝑥 𝑑 𝑦 𝑑 𝑥 + 6 𝑒 𝑦 𝑑 𝑦 𝑑 𝑥 = 0 y+x dx dy ​ +6e y dx dy ​ =0 From here: 𝑦 ′ ( 𝑥 + 6 𝑒 𝑦 ) = − 𝑦 y ′ (x+6e y )=−y Substituting 𝑥 = 0 x=0 and 𝑦 = 1 y=1: 𝑦 ′ ( 0 + 6 𝑒 ) = − 1 y ′ (0+6e)=−1 𝑦 ′ = − 1 6 𝑒 y ′ =− 6e 1 ​ Question 3 Solve the differential equations: 𝑥 𝑦 ′ − 2 𝑦 = 𝑥 3 𝑒 𝑥 xy ′ −2y=x 3 e x ( 2 𝑦   𝑑 𝑥 + 𝑑 𝑦 ) 𝑒 2 𝑥 = 0 (2ydx+dy)e 2x =0 Answer 1. 𝑥 𝑦 ′ − 2 𝑦 = 𝑥 3 𝑒 𝑥 xy ′ −2y=x 3 e x The integrating factor 𝑒 ∫ − 2 / 𝑥   𝑑 𝑥 = 1 𝑥 2 e ∫−2/xdx = x 2 1 ​ : 𝑑 𝑑 𝑥 ( 𝑦 / 𝑥 2 ) = 𝑒 𝑥 dx d ​ (y/x 2 )=e x 𝑦 = 𝑥 2 𝑒 𝑥 ∫ 𝑒 𝑥   𝑑 𝑥 + 𝑐 𝑥 2 y=x 2 e x ∫e x dx+cx 2 𝑦 = 𝑥 2 𝑒 𝑥 ( 𝑒 𝑥 ) + 𝑐 𝑥 2 y=x 2 e x (e x )+cx 2 𝑦 = 𝑥 2 𝑒 2 𝑥 + 𝑐 𝑥 2 y=x 2 e 2x +cx 2 ( 2 𝑦   𝑑 𝑥 + 𝑑 𝑦 ) 𝑒 2 𝑥 = 0 (2ydx+dy)e 2x =0 𝑑 𝑦 𝑦 = − 2   𝑑 𝑥 y dy ​ =−2dx ln ⁡ 𝑦 = − 2 𝑥 + 𝑐 lny=−2x+c 𝑦 = 𝑒 − 2 𝑥 + 𝑐 y=e −2x+c 𝑦 = 𝑒 − 2 𝑥 + 𝑐 y=e −2x +c Question 4 Find the solution of 𝑦 ( 𝑡 ) = 𝑘 𝑦 ( 𝑡 ) y(t)=ky(t) with 𝑦 ( 1 ) = 5 𝑦 ( 1 ) = 4 ,   𝑦 = 𝐶 𝑒 𝑘 𝑡 y(1)=5y(1)=4,y=Ce kt . Answer Write: 𝑑 𝑦 𝑑 𝑡 = 𝑘 𝑦 dt dy ​ =ky ln ⁡ 𝑦 = 𝑘 𝑡 + 𝑐 lny=kt+c 𝑦 = 𝑒 ( 𝑘 𝑡 + 𝑐 ) = 𝑒 𝑐 𝑒 𝑘 𝑡 y=e (kt+c) =e c e kt Let 𝐶 = 𝑒 𝑐 C=e c , so 𝑦 = 𝐶 𝑒 𝑘 𝑡 y=Ce kt : 𝑘 = ln ⁡ ( 5 ) − ln ⁡ ( 1 ) 𝑡 = 0.8 k= t ln(5)−ln(1) ​ =0.8 Thus, 𝑦 = 5 𝑒 0.8 𝑡 y=5e 0.8t . Question 5 The integrating factor method for second-order equations: (a) Show that 𝜇 ′ ′ − 𝑝 ( 𝑡 ) 𝜇 ′ + 𝑞 ( 𝑡 ) 𝜇 = 0 μ ′′ −p(t)μ ′ +q(t)μ=0. Answer From differentiating 𝜇 ( 𝑡 ) ( 𝑦 ′ + 𝑦 ) = 𝜇 ( 𝑡 ) 𝑓 ( 𝑡 ) μ(t)(y ′ +y)=μ(t)f(t): 𝜇 ( 𝑡 ) 𝑦 ′ + 𝜇 ′ ( 𝑡 ) 𝑦 + 𝑔 ( 𝑡 ) = 𝜇 ( 𝑡 ) 𝑓 ( 𝑡 ) μ(t)y ′ +μ ′ (t)y+g(t)=μ(t)f(t) Let 𝑔 ( 𝑡 ) = 𝜇 ( 𝑡 ) 𝑝 ( 𝑡 ) + 𝜇 ′ ( 𝑡 ) g(t)=μ(t)p(t)+μ ′ (t): 𝜇 ′ + ∫ 𝑔 ( 𝑡 ) = 𝜇 ′ μ ′ +∫g(t)=μ ′ From the adjoint: 𝜇 ′ ′ − 𝑝 ( 𝑡 ) 𝜇 ′ + 𝑞 ( 𝑡 ) 𝜇 = 0 μ ′′ −p(t)μ ′ +q(t)μ=0.