Question

Y is equal to a cos 3x + b sin 3x then find d squarey upon DX square​

Answer 1

d²y/dx² = - 9y

Step-by-step explanation :

Given, y = a cos3x + b sin3x ..... (1)

Now differentiating both sides of (1) no. equation with respext to x, we get

dy/dx = - 3a sin3x + 3b cos3x ..... (2)

Again differentiating both sides of (2) no. equation with respect to x, we have

d²y/dx^2 = - 9a cos3x - 9b sin3x

= - 9 (a cos3x + b sin3x)

= - 9y, by (1)

Therefore, d²y/dx² = - 9y

Related question :

if ax^2+2hxy+by^2=1 then (d^2y)/(dx^2)=(h^2-ab)/(hx+by)^3 - https://brainly.in/question/15904397

Answer 2

$\mathbf{\frac{\mathrm{d^{2}}Y }{\mathrm{d} x^{2}}=-9Y}$

Step-by-step explanation:

• We know that

        $\mathbf{\frac{\mathrm{d} \sin 3x }{\mathrm{d} x }=3\cos 3x }$

        and

        $\mathbf{\frac{\mathrm{d} \cos 3x }{\mathrm{d} x }=-3\sin 3x }$

• Here it is given that

        $\mathbf{Y=a\cos 3x+b\sin 3x}$      ...1)

• On differentiating above equation with respect to x, we get

        $\mathbf{\frac{\mathrm{d}Y }{\mathrm{d} x}=a\frac{\mathrm{d} \cos 3x}{\mathrm{d} x}+b\frac{\mathrm{d} \sin 3x}{\mathrm{d} x}}$

        So it be can written as

        $\mathbf{\frac{\mathrm{d}Y }{\mathrm{d} x}=a(-3\sin 3x)+b(3\cos 3x)}$

        $\mathbf{\frac{\mathrm{d}Y }{\mathrm{d} x}=-3a\sin 3x+3b\cos 3x}$       ...2)

• Again on differentiating above equation with respect to x, we get

        $\mathbf{\frac{\mathrm{d^{2}}Y }{\mathrm{d} x^{2}}=-3a\frac{\mathrm{d} \sin 3x}{\mathrm{d} x}+3b\frac{\mathrm{d} \cos 3x}{\mathrm{d} x}}$

        $\mathbf{\frac{\mathrm{d^{2}}Y }{\mathrm{d} x^{2}}=-3a(3\cos 3x)+3b(-3\sin 3x)}$

        $\mathbf{\frac{\mathrm{d^{2}}Y }{\mathrm{d} x^{2}}=-9a\cos 3x-9b\sin 3x}$

        $\mathbf{\frac{\mathrm{d^{2}}Y }{\mathrm{d} x^{2}}=-9(a\cos 3x+b\sin 3x)}$     ...3)

• From equation 1) and equation 3), we can write above equation

        $\mathbf{\frac{\mathrm{d^{2}}Y }{\mathrm{d} x^{2}}=-9Y}$