Solve the equation:y″+2y′+2y=0, y(π4)=2, y′(π4)=−2

Name: Solve the equation:y″+2y′+2y=0, y(π4)=2, y′(π4)=−2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the equation:y″+2y′+2y=0, y(π4)=2, y′(π4)=−2

Question

Solve the equation:y″+2y′+2y=0, y(π4)=2, y′(π4)=−2

uqhekekocj8f · Accepted Answer

Given that:The second order differential equationy″+2y′+2y=0The initial condition y(π4)=2 and y′(π4)=−2Use:A second order linear, homogeneous ordinary differential equation has the form ofy″+py′+qy=0The characteristic equation : r2+pr+q=0 to the general solution with two complex root r1≠qr2where r1=α+i⋅β,r2=α−i⋅βThe general solution take the form e=eα⋅t(c1cos⁡(β⋅t)+c2sin⁡(β⋅t))The characteristic equation of the given second order , linear , homogeneous ordinary differential equation isr″+2r′+2=0This is the quadratic polynomial of the form ax2+bx+c=0To use the quadratic formula to find the root of the above polynomial. x=−b±b2−4ac2aComparing the given polynomial with ax2+bx+c=0To geta=1, b=2, c=2Plug the values of a=1, b=2, c=2 in the quadratic formula.To get,r=−2±22−4(1)(2)2(1)=−2±4−82=2±−42=2±i22=2(−1±i)2=−1±iTo getr=−1±iThe root of the given polynomial are r1=−1+i, r2=−1−iThen,The general solution of the given differential equation:y=eα⋅t(c1cos⁡(β⋅t)+c2sin⁡(β⋅t))Plug the values of a=−1 and β=1To get,y(t)=e−t(c1cos⁡t+c2sin⁡t)To find y&#039;(t)Differentiate the above equation with respect to t,To gety′(t)=−e−t(c1cos⁡t+c2sin⁡t)+e−t(−c1sin⁡t+c2cos⁡t)Use the initial condition y(π4)=2, and y′(π4)=−2Plug t=π4 in y(t)To get,y(π4)=e−π4(c1cos⁡(π4)+c2sin⁡(π4))2=e−π4(c1(12)+c2(12))2=e−π4c12+e−π4c22Plug the t=π4 in y&#039;(t)To gety′(π4)=−e−π4(c1cos⁡(π4)+c2sin⁡(π4))+e−π4(−c1sin⁡(π4)+c2cos⁡(π4))−2=−e−π4(c1(12)+c2(12))+e−π4(−c1(12)+c2(12))−2=−e−π4c12−e−π4c22−e−π4c12+e−π4c22−2=−e−π4c12Multiplying on both side by 2−e−π4To get,−22−e−π4=c1To getc1=22e−π4Plug the value of c1=(22)e−π4 in the equation 2=e−π4(c1(12)+c2(12))2=c1e−π42+c2e−π422=(22e−π4)e−π42+c2e−π422=22e−π4e−π42+c2e−π422=2+c2e−π42Substract 2 on both side,To get0=c2e−π42c2e−π4=0Divide on both side be e−π4to getc2=0Plug the value of c1=22e−π4 and c2=0 in the general solution of the differential equationy(t)=e−t(22e−π4cos⁡t+0sin⁡t)To get,y(t)=22e−te−π4Therefore, The general solution of the given differential with the given initial condition y(π4)=2 and y′(π4)=−2 isy(t)=22e−te−π4

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Solve the equation: \(\displaystyle{y}{''}+{2}{y}'+{2}{y}={0},\ {y}{\left({\frac{{\pi}}{{{4}}}}\right)}={2},\ {y}'{\left({\frac{{\pi}}{{{4}}}}\right)}=-{2}\)

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