Thread: 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!

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)
Then add your equations,
[(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.

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