Find the solution of the following Differential Equations y′−4y=2x−4x2

Name: Find the solution of the following Differential Equations y′−4y=2x−4x2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the solution of the following Differential Equations y′−4y=2x−4x2

Question

censoratojk · Accepted Answer

Find the solution of the following DEy′−4y=2x−4x2Complete solution =CF+PICF=y&#039;-4y=0(D−4)y=0m−4=0m=4CF=C1e4xPI=af(D)⇒2x−4x2D−4PI=1−4+D(2x−4x2)=1−4(1−D4)(2x−4x2)−4+D=−4(1−D4)=−14(1−D4)−1(2x−4x2)(1−D)−1=1+D+D2+D3+…(1+D)−1=1−D+D2−D3+…D(2x−4x2)=ddx(2x−4x2)=2−8xD2(2x−4x2)=d2dx2(2x−4x2)=−8D3(2x−4x2)=d3dx3(2x−4x2)=0PI=−14(1+D4+(D4)2+(D4)3+…)(2x−4x2)

jean2098 · Accepted Answer

y′−4y=2x−4x2 (1)Clearly, the given differential equation is of the form y′+py=aThus, differential eq (1) is linear, where P=−4 and a=2x−4x2Now, integrating factor, I.F=e∫pdx=e∫dx⇒I.F=e−4xThe solution of given differential equation (1) is given by,y(IF)=∫q(IF)dx+c, c is a constant⇒y.e−4x=∫(2x−4x2)e−4xdx+c (2)Now ∫(2x−4x2)e−4xdx=∫euu(u+2)16du⇒∫(2x−4x2)edx−4x∫euu(u+2)du (3)116∫euu(u+2)du=116[(u+2)(euu−eu)−∫(euu−eu)du]=116[(u+2)(euu−eu)−euu+2eu]=116[u2eu−ueu+2ueu−2eu−ueu+2eu]=116u2euHence, from eq (3), we get∫(2x−4x2)e−4xx=116u2euu=−4x, we get

karton · Accepted Answer

⇒y.e−4x=∫(2x−4x2)e−4xdx+c(2)∫(2x−4x2)e−4xdx=∫euu(u+2)16du⇒∫(2x−4x2)edx−4x∫euu(u+2)du(3)116∫euu(u+2)du=116[(u+2)(euu−eu)−∫(euu−eu)du]=116[(u+2)(euu−eu)−euu+2eu]=116[u2eu−ueu+2ueu−2eu−ueu+2eu]=116u2euHence, from eq (3), we getint(2x−4x2)e−4xx=116u2euu=−4x, we get∫(2x−4x2)e−4xdx=116×(−4x)2e−4x=x2e−4xHence, from eq (2), we gety.e−4x=x2e−4x+c⇒y=x2+ce4xAnswer: y=x2+ce4x