d square y divided by dt square+2 dy divided by dx_3y=sint, given that y=dy divided by dx=0 when t=0
d(dy/dt)/dt+2dy/dx-3y=sin(t); y=dy/dx=0 when t=0.
y=f(t); x=g(t); dy/dt=f'; dx/dt=g'; d(dy/dt)/dt=f"; dy/dx * dx/dt=dy/dt, g'dy/dx=f'; dy/dx=f'/g'.
f"+2f'/g'-3f=sin(t) or g'f"+2f'-3fg'=g'sin(t).
f(0)=0; f'(0)/g'(0)=0, so f'(0)=0 (because dy/dx=0=f'(0)/g'(0)).
Also, g'(0) cannot be zero, otherwise dy/dx would be undefinable at t=0.
But for f'/g', we would have an expression on the left that could be solved as an auxiliary, or characteristic, equation and we could apply the homogeneous solution fH=Ae^-3t + Be^t where fH denotes the homogeneous component of the solution, and A and B are constants and the exponents are simply the solutions of m^2+2m-3=(m+3)(m-1)=0 giving us roots -3 and 1. If g'=1 then clearly g' is non-zero, which is a requirement. But if g'=1 x=g(t)=t+C where C is a constant and fH=Ae^-3t + Be^t. xH=gH=t, where H denotes the homogeneous component. Could this be a solution? This is the homogeneous component of the solution fH, where complete f=fH+fP, where fP denotes the particular solution.
Now we need to resolve the right-hand side. This is the particular solution fP and we can assume that it is of the form fP=Csin(t)+Dcost(t).
Applying the derivatives we have: Ccos(t)-Dsin(t) as 1st, -Csin(t)-Dcos(t) as 2nd. These must satisfy the original DE.
sin(t): -C-2D/g'-3C=1; cos(t): -D+2C/g'-3D=0. 4C+2D/g'+1=0 and 2C/g'=4D, so C=2Dg'. 8Dg'+2D/g'+1=0, 8Dg'^2+2D+g'=0. 2D(4g'^2+1)=-g'. This gives us fP=-sin(t)/5-g'cos(t)/(8g'^2+2).
Combining the auxiliary and particular components, y=f(t)=fH+fP=Ae^-3t + Be^t-sin(t)/5-g'cos(t)/(8g'^2+2). If we can replace g' with 1 then D=-1/10, C=-1/5, y=Ae^-3t + Be^t-sin(t)/5-cos(t)/10 and x=t. The initial conditions have f(0)=0=f'(0), so 0=A+B-1/10 and -3A+B-1/5=0. From these, -3A+1/10-A-1/5=0, -4A-1/10=0, A=-1/40, 0=-1/40+B-1/10, making B=1/8.
Thus, y=(e^t)/8-(e^-3t)/40-sin(t)/5-cos(t)/10, x=t or y=(e^x)/8-(e^-3x)/40-sin(x)/5-cos(x)/10.
y'=(e^t)/8+3e^(-3t)/40-cos(t)/5+sin(t)/10; x'=g'=1, y'=f', dy/dx=dy/dt=f';
(e^t)/8-9e^(-3t)/40+sin(t)/5+cos(t)/10 [this is y"]
+(e^t)/4+3e^(-3t)/20-2cos(t)/5+sin(t)/5 [this is 2dy/dx=2y']
-3(e^t)/8+3(e^-3t)/40+3sin(t)/5+3cos(t)/10= [this is -3y]
[ADDENDUM: Why the auxiliary, or characteristic, equation works.
Consider the equation y=Ae^ax+Be^bx, where a and b are constants.
Now consider a 2nd degree DE: y"+Py'+Qy=0, where P and Q are constants. We can replace the derivatives:
a^2Ae^ax+b^2Be^bx+PaAe^ax+PbBe^bx+QAe^ax+QBe^bx=0, which can be written:
Since the exponential components cannot be <0, the quadratics in a and b must each be zero and they must be the roots of the same equation: e.g., z^2+Pz+Q=0 where the roots are z=a and b, with P and Q as the coefficients. In the question, P=2 and Q=-3 so z^2+2z-3=0 is the quadratic, so (z+3)(z-1)=0 and a=-3 and b=1, so fH=Ae^-3t+Be^t.] Read More: ...