# Thread: Differential Equation Proof

1. ## Differential Equation Proof

y1(x) and y2(x) are solutions of

y'' + P(x)*y' + Q(x)*y = R1(x)

and

y'' + P(x)*y' + Q(x)*y = R2(x)

respectively. I need to show that y(x) = y1(x) + y2(x) is a solution of

y'' + P(x)*y' + Q(x)*y = R1(x) + R2(x).

(y1(x), y2(x), R1(x), and R2(x) should be read as "y 'sub' one of x", and should be understood as referring to separate solutions and equations.)

I took this class to finish up a math minor, but I was never good with this sort of thing. I'm into numbers, not theory and symbols.

Any help is appreciated!

2. Originally Posted by primasapere
y1(x) and y2(x) are solutions of

y'' + P(x)*y' + Q(x)*y = R1(x)

and

y'' + P(x)*y' + Q(x)*y = R2(x)

respectively. I need to show that y(x) = y1(x) + y2(x) is a solution of

y'' + P(x)*y' + Q(x)*y = R1(x) + R2(x).
Okay,
You know that,
(y1)''+p(x)(y1)'+q(x)(y1)=r1(x)
(y2)''+p(x)(y2)'+q(x)(y2)=r2(x)
[(y1)''+(y2)'']+[p(x)(y1)'+p(x)(y1)']+[q(x)(y1)+q(x)(y2)]=r1(x)+r2(x)
Factor,
[(y1)''+(y2)'']+p(x)[(y1)'+(y2)']+q(x)[(y1)+(y2)]=r1(x)+r2(x)
Use the addition property of the derivative,
(Note if you want to sound cool say diffrenciation is a homomorphism)
(y1+y2)''+p(x)(y1+y2)'+q(x)(y1+y2)=r1(x)+r2(x)

Therefore we see that,
y=y1+y2
Is the solution.

3. You, sir, are amazing. Thank you so much.