Question

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

A. y=3sinx
B. y=3xsinx-5xcosx
C. y=-3/2xcosx
D. y=3cosx
E. y=3/2xsinx

Answer 1

Answer:

C

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$. Differentiate again to get $y''=(3\cos x)'=-3\sin x$, then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$, then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$. Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D.$y'=-3\sin x$ and $y''=-3\cos x$, then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$. Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

Answer 2

Double differentiation of a function is the rate of change of the rate of change of one variable with respect to other.

The solution of the given equation is,

$y=-\dfrac{3}{2} x\cos x$

Thus option C is the correct option.

Given information-

$y"+y=3 \sin x$

What is double differentiation?

Double differentiation of a function is the rate of change of the rate of change of one variable with respect to other.

As the given function is the double differentiate function. To find the function we need double differentiate all the options.

• A) The function given in the option A is,

$y=3sinx$

Differentiate above function with respect to the x,

$y'=3cosx$

Differentiate above function again with respect to the x,

$y"=-3sinx$

Which is not equal to the given expression. This is not the correct option.

• B) The function given in the option B is,

$y=3\sin x-5x\cos x$

Differentiate above function with respect to the x,

$y'=3\cos x+5x\sin x-5\cos x$

Differentiate above function again with respect to the x,

$y"=-3\sin x+5x\sin x+5x\cos x+5\sin x\\y"=2\sin x+5x(\sin x+\cos x)$

Which is not equal to the given expression. This is not the correct option.

• C) The function given in the option C is,

$y=-\dfrac{3}{2} x\cos x$

Differentiate above function with respect to the x,

$y'=-\dfrac{3}{2}(- x\sin x+cosx)$

Differentiate above function again with respect to the x,

$y"=-\dfrac{3}{2}(-\sin x-x\cos x+(-\sin x))\\y"=\dfrac{3}{2}(2\sin x-x\cos x)\\y"=3\sin x-\dfrac{3}{2} x\cos x\\y"+y=3\sin x-\dfrac{3}{2} x\cos x+\dfrac{3}{2} x\cos x\\\\y"+y=3\sin x$

Which is  equal to the given expression. This is the correct option.

• D) The function given in the option D is,

$y=3\cos x$

Differentiate above function with respect to the x,

$y'=-3\sin x$

Differentiate above function again with respect to the x,

$y"=-3\cos x$

Which is not equal to the given expression. This is not the correct option.

• E) The function given in the option E is,

$y=\dfrac{3}{2} xsinx$

Differentiate above function with respect to the x,

$y'=\dfrac{3}{2} (\sin x +x\cos x)$

Differentiate above function again with respect to the x,

$y"=\dfrac{3}{2} (-\cos x +\cos x-x\sin x)\\y"=-x\sin x$

Which is not equal to the given expression. This is not the correct option.

Hence the solution of the given equation is,

$y=-\dfrac{3}{2} x\cos x$

Thus option C is the correct option.

Learn more about the double differentiation here;

https://brainly.com/question/12381251