MATB41H3 Lecture : solutions5(1)
DIFFERENTIAL EQUATIONS I MATB44 ASSIGNMENT 5. SOLUTION.
1. 1. Solve
y′′ −3y′= 2e2xsin x
The characteristic equation
r2−3r= 0 ⇒
r1= 0, r2= 3.
Here is general solution of the homogeneous equation
y=c1y1+c2y2=c1+c2e3x.
To find a solution for the non-homogeneous equation we set
y=Ae2xcos x+Be2xsin x
Put it in the LHS
LHS = (Ae2xcos x+Be2xsin x)′′ −3(Ae2xcos x+Bte2xsin x)′
= (4Ae2xcos x−4Ae2xsin x−Ae2xcos x+ 4Be2xsin x+ 4Be2xcos x−Be2xsin x)
−3(2Ae2xcos x−Ae2xsin x+ 2Be2xsin x+Be2xcos x)
= (−3A+B)e2xcos x+ (−A−3B)e2xsin x
Thus,
(−3A+B)e2xcos x+ (−A−3B)e2xsin x= 2e2xsin x⇒
{−3A+B= 0
−A−3B= 2
⇒A=−1
5, B =−3
5
So,
y0=−1
5e2xcos x−3
5e2xsin x.
Here is general solution of the equation
y=c1y1+c2y2+y0=c1+c2e3x
=c1+c2e3x−1
5e2xcos x−3
5e2xsin x.
2. Solve
y′′ + 4y= sin 2x
r2+ 4 = 0 ⇒
r1,2=±2i.
Here is general solution of the homogeneous equation
y=c1y1+c2y2=c1cos 2x+c2sin 2x.
For that matter the non-homogeneous equation has no solutions of the shape
y=Acos 2x+Bsin 2x.
1
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Document Summary
The characteristic equation y 3y = 2e2x sin x r2 3r = 0 r2 = 3. r1 = 0, To nd a solution for the non-homogeneous equation we set y = ae2x cos x + be2x sin x y = c1y1 + c2y2 = c1 + c2e3x. Lhs = (ae2x cos x + be2x sin x) 3(ae2x cos x + bte2x sin x) . = (4ae2x cos x 4ae2x sin x ae2x cos x + 4be2x sin x + 4be2x cos x be2x sin x) 3(2ae2x cos x ae2x sin x + 2be2x sin x + be2x cos x) = ( 3a + b)e2x cos x + ( a 3b)e2x sin x ( 3a + b)e2x cos x + ( a 3b)e2x sin x = 2e2x sin x . Here is general solution of the equation y = c1y1 + c2y2 + y0 = c1 + c2e3x.
