Y" + 6y' + 9y = 1 + x First, you've got acquired to remedy the equation without 2d member (with out right part). Y" + 6y' + 9y = zero Accordig this equation, the solution is structure : y = e^rx So, you can write : y' = r.E^rx and : y'' = r².E^rx attribute equation with out 2d member : r² + 6r + 9 = zero Polynomial like : ax² + bx + c, where : a = 1 b = 6 c = 9 Δ = b² - 4ac (discriminant) Δ = 6² - four(1 * 9) = 36 - 36 = 0 r = - b / 2a = - 6 / 2 = - three you can conclude that : y = e^3x Then, you need to discover a specified resolution of the equation with the second member you already know that answer is layout : y = ax + b y' = a y'' = 0 Now, exchange y'', y' and y by them values in the equation with second member. Y" + 6y' + 9y = 1 + x zero + 6a + 9(ax + b) = 1 + x 6a + 9ax + 9b = 1 + x [9a]x + [6a + 9b] = x + 1 compare the 2 facets, and you will find : 9a = 1, thus a = 1/9 6a + 9b = 1, thus : b = (1 - 6a)/9 = [1 - (6/9)]/9 = (three/9)/9 = (1/three)/9 = 1/27 Now, replace a and b by using them values in y = ax + b and also you get : y = (x/9) + (1/27) resolution = (answer without 2d member) + (specific resolution with 2nd member) y = e^3x + (x/9) + (1/27) i am French, sorry for language.
Answers & Comments
Verified answer
for particular solution , y = A cos3x + B sin3x , and plug in y'' +y=sin3x to find A , B
Y" + 6y' + 9y = 1 + x First, you've got acquired to remedy the equation without 2d member (with out right part). Y" + 6y' + 9y = zero Accordig this equation, the solution is structure : y = e^rx So, you can write : y' = r.E^rx and : y'' = r².E^rx attribute equation with out 2d member : r² + 6r + 9 = zero Polynomial like : ax² + bx + c, where : a = 1 b = 6 c = 9 Δ = b² - 4ac (discriminant) Δ = 6² - four(1 * 9) = 36 - 36 = 0 r = - b / 2a = - 6 / 2 = - three you can conclude that : y = e^3x Then, you need to discover a specified resolution of the equation with the second member you already know that answer is layout : y = ax + b y' = a y'' = 0 Now, exchange y'', y' and y by them values in the equation with second member. Y" + 6y' + 9y = 1 + x zero + 6a + 9(ax + b) = 1 + x 6a + 9ax + 9b = 1 + x [9a]x + [6a + 9b] = x + 1 compare the 2 facets, and you will find : 9a = 1, thus a = 1/9 6a + 9b = 1, thus : b = (1 - 6a)/9 = [1 - (6/9)]/9 = (three/9)/9 = (1/three)/9 = 1/27 Now, replace a and b by using them values in y = ax + b and also you get : y = (x/9) + (1/27) resolution = (answer without 2d member) + (specific resolution with 2nd member) y = e^3x + (x/9) + (1/27) i am French, sorry for language.
y(hom)=Acosx+Bsinx
y(particular)=Kcos3x+Lsin3x
y(p)'=-3Ksin3x+3Lcos3x
y''(p)=-9Kcos3x-9Lsin3x
putting these values in the ODE we have L=1/8 and K=0
so y(particular)=(1/8)sin3x
y(general)=Acosx+Bsinx+(1/8)sin3x