The function that is the solution of the differential y" + y = sin(x) is: y(x) = -1/2(x cos x)
A function is an expression, rule, or law in mathematics that describes a connection between one factor (the independent variable) and another variable (the dependent variable).
Take a look at the the following differential equation:
y" + y = sin (x)
The auxiiliary equation is
m² + 1 = 0
m² + 1-1 = -1
m² + 0 = -1
m² = -1
m = ±√-1
m = ±i
So, the complimentary function is yₐ (x) = c₁ cos x + c₂ sin x
Let the particular integral be:
yₙ (x) = A cos x + B sin x
yₙ '(x) = - A sin x + B cos x
yₙ ''(x) = - A cos x + B cos x
yₙ ''(x) = - (A cos x + B cos x)
After we have substituted yₙ (x); and yₙ''(x) in the given differential equation
y'' + y = sin (x)
= - (A cos x + B cos x) + (A cos x + B cos x) = sin(x)
0 = sin (x)
If we take the particular integral to be:
yₙ (x) = x(A cos x + B cos x)
yₙ '(x) = x(-A sin x + B cos x) + A cos x + B cos x
yₙ ''(x) = x(-A cos x - B sin x) - A sin x + B cos x + -A sin x + B cos x
Substitute yₙ (x), yₙ''(x) into the stated differential equation
y'' + y = sin (x)
x (-Acosx - Bsin x) - Asinx + Bcosx + (-Asinx + Bcosx) - x (Acosx + Bsin x) = sin (x)
-Axcosx - Bxsin x - Asinx + Bcosx -Asinx + Bcosx - Axcosx + Bxsin x = sin (x)
-2Asinx + 2Bcosx = sin(x)
Compare the coefficients of like terms on both sides of the equation
-2A = 1, B = 0
A = -1/2, B = 0
Substitute A = -1/2, B =0 into the assumed solution.
yₙ(x) = x((-1/2)cosx + (0) sinx)
= -1/2xcosx +0
= -(1/2)xcosx
Now, the general solution for the given differential equation is:
y(x) = yₓ(x) +yₙ (x)
y (x) - c₁cosx + c₂sin x -1/2x cosx
Hence, the solution is:
y(x) = -1/2xcosx
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Full Question:
Which of the following functions are solutions of the differential equation y'' + y = sin x? (Select all that apply.)
A) y = 1 2 x sin x
B) y = cos x
C) y = x sin x − 5x cos x
D) y = − 1 /2 x cos x
E) y = sin x
