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In this exercise you will solve the initial value problem y'' - 20y' + 100y = e^{-10x}/(1 + x^2), y(0) = 5, y'(0) = -5. (1) Let C1 and C2 be arbitrary constants. The general solution to the related homogeneous differential equation y'' - 20y' + 100y = 0 is the function yh(x) = C1 y1(x) + C2 y2(x) = C1 [ ] + C2 [ ]. NOTE: The order in which you enter the answers is important; that is, C1f(x) + C2g(x) ? C1g(x) + C2f(x). (2) The particular solution yp(x) to the differential equation y'' + 20y' + 100y = e^{-10x}/(1 + x^2) is of the form yp(x) = y1(x) u1(x) + y2(x) u2(x) where u'1(x) = [ ] and u'2(x) = [ ]. (3) The most general solution to the non-homogeneous differential equation y'' - 20y' + 100y = e^{-10x}/(1 + x^2) is y = [ ] + [ ] ? [ ] / [ ] [ ] dt + [ ] ? [ ] / [ ]

          In this exercise you will solve the initial value problem

y'' - 20y' + 100y = e^{-10x}/(1 + x^2), y(0) = 5, y'(0) = -5.

(1) Let C1 and C2 be arbitrary constants. The general solution to the related homogeneous differential equation y'' - 20y' + 100y = 0 is the function yh(x) = C1 y1(x) + C2 y2(x) = C1 [ ] + C2 [ ].

NOTE: The order in which you enter the answers is important; that is, C1f(x) + C2g(x) ? C1g(x) + C2f(x).

(2) The particular solution yp(x) to the differential equation y'' + 20y' + 100y = e^{-10x}/(1 + x^2) is of the form yp(x) = y1(x) u1(x) + y2(x) u2(x) where u'1(x) = [ ] and u'2(x) = [ ].

(3) The most general solution to the non-homogeneous differential equation y'' - 20y' + 100y = e^{-10x}/(1 + x^2) is

y = [ ] + [ ] ? [ ] / [ ] [ ] dt + [ ] ? [ ] / [ ]

Show more…

In this exercise you will solve the initial value problem

y” - 20y' + 100y = e^-10x/(1 + x^2), y(0) = 5, y'(0) = -5.

(1) Let C1 and C2 be arbitrary constants. The general solution to the related homogeneous differential equation y” - 20y' + 100y = 0 is the function yh(x) = C1 y1(x) + C2 y2(x) = C1 [ ] + C2 [ ].

NOTE: The order in which you enter the answers is important; that is, C1f(x) + C2g(x) ? C1g(x) + C2f(x).

(2) The particular solution yp(x) to the differential equation y” + 20y' + 100y = e^-10x/(1 + x^2) is of the form yp(x) = y1(x) u1(x) + y2(x) u2(x) where u'1(x) = [ ] and u'2(x) = [ ].

(3) The most general solution to the non-homogeneous differential equation y” - 20y' + 100y = e^-10x/(1 + x^2) is

y = [ ] + [ ] ? [ ] / [ ] [ ] dt + [ ] ? [ ] / [ ]

Added by Xavier C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

11/20/2022

Step 1

This is a second order linear homogeneous differential equation with constant coefficients. The characteristic equation of this differential equation is r^2 + 20r + 10 = 0. Solving this quadratic equation, we get r = -10 ± √(100 - 40), which gives r = -10 ± √60. Show more…

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In this exercise you will solve the initial value problem y'' - 20y' + 100y = e^{-10x}/(1 + x^2), y(0) = 5, y'(0) = -5. (1) Let C1 and C2 be arbitrary constants. The general solution to the related homogeneous differential equation y'' - 20y' + 100y = 0 is the function yh(x) = C1 y1(x) + C2 y2(x) = C1 + C2 . NOTE: The order in which you enter the answers is important; that is, C1f(x) + C2g(x) != C1g(x) + C2f(x). (2) The particular solution yp(x) to the differential equation y'' + 20y' + 100y = e^{-10x}/(1 + x^2) is of the form yp(x) = y1(x) u1(x) + y2(x) u2(x) where u'1(x) = and u'2(x) = . (3) The most general solution to the non-homogeneous differential equation y'' - 20y' + 100y = e^{-10x}/(1 + x^2) is y = + dt +

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00:01 So, here we are given the equation which is y double dash minus 20, y dash plus 100 of y is equal to, this is equals to e raised to the power minus 10 of x divided by a 1 plus x square, where y of 0 is equal to 5 and y dash of 0 is equals to minus 5.

00:21 So we have to find out the solution of this equation.

00:24 So the auxiliary form of the equation, auxiliary form will be r square minus 20 r plus 100 is equal to 0.

00:38 So we can write it as r minus 10 whole square is equal to 0.

00:42 So from here we get the roots of the equation 10 and 10.

00:46 So we can say that the y n of x is equal to c1, e raised to the power 10 of x plus c2 x, e raise to the power 10 of x.

00:55 So this is the solution of this equation.

01:00 Now in the second part, we have to find out the value of where we are given that y is equal to e1, e10 raised to the power x.

01:11 And y2 is equal to x, e raised to the power 10 of x.

01:14 So here we have to find it's general assumption.

01:17 So yp is equal to y1 x, u1, x plus y2x, y2x, plus y2 x, u2x, u2x, u2, u.

01:29 U2x.

01:30 So from here we can say that u1 dash is equals to minus y2, k of x divided by the w of x and u2 dash will be y1, r of x divided by the w of x.

01:45 Let's say this is the equation number 1.

01:47 So from here we can say that r of x is right hand side of given equation...

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