How to solve y=x^3 sin^2 5x?
y = x^3 sin^2 (5x) By product rule dy/dx = x^3 . d/dx [sin^2(5x)] + sin^2(5x) . d/dx (x^3) = x^3 . 2 sin(5x) cos(5x) (5) + sin^2(5x) . (3x^2) = x^3 . 10sin(5x) cos(5x) + 3x^2 sin^2(5x) = 10x^3 sin(5x) cos(5x) + 3x^2 sin^2(5x)
How do I solve 3cos^2x - 2root 3 sinx cosx -3 sin^2x = 0?
First of all we simplify the given equation.Simplification of the equation:[math]3\cos^2x-3\sin^2x-\sqrt3\;2\sin x\,\cos x=0[/math][math]3(\cos^2x-\sin^2x)-\sqrt3\;(2\sin x\,\cos x)=0[/math][Trigonometric identities][math]\because\;\cos^2x-\sin^2x=\cos2x[/math][math]2\,\sin x\;\cos x=\sin 2x[/math][math]3(\cos\,2x)-\sqrt3(\sin\,2x)=0[/math][math]3(\cos\,2x)=\sqrt3\;\sin\,2x[/math][math]\dfrac{3}{\sqrt3}=\dfrac{\sin\,2x}{\cos\,2x}[/math][math]\sqrt3=\tan\,2x[/math][math]\tan^{-1}(\sqrt3)=2x[/math](According to the property of inverse trigonometric functions)[math]\tan^{-1}\left(\tan\dfrac{\pi}{3}\right)=2x[/math][math]\dfrac{\pi}{3}=2x[/math][math]\dfrac{\pi}{6}=x[/math]It is the solution of this problem.
What is dy/dx of sin^2 (x)?
This is the Answer of your question.The answer is sin2xThe picture has all the detailed steps and the formulas which are used to solve this problem.Thank You.
Y''-4y = (x^2-3)sin(2x) where y(0)=0 , y'(0)=1?
Try using: Yp = (Ax^2 + Bx + c)cos(2x) + (Dx^2 + Ex + F)sin2x For undetermined coefficients method. Take 2 derivatives and plug into the differential equation. Then solve for coefficients A-F. Your solution y(x) will then be yh + yp, where yh is the homogenous portion. That will be your general solution. Plug in initial conditions.
How to solve y" + 4y = 3 sin 2x by undetermined coefficients?
We first find the general solution to the homogeneous system. i.e. y" + 4y = 0 This leads to y(h) = Acos2x + Bsin2x Now as the right side of the non-homogeneous system....3sin2x is contained within the general solution we try y(p) = Axsin2x + Bxcos2x => y'(p) = Asin2x + 2Axcos2x + Bcos2x - 2Bxsin2x i.e. y"(p) = 2Acos2x + 2Acos2x - 4Axsin2x - 2Bsin2x - 2Bsin2x - 4Bxcos2x Then, y"(p) + 4y(p) = 4Acos2x - 4Bsin2x => 3sin2x so, A = 0 and B = -3/4 Therefore, y(p) = -(3x/4)cos2x....so, putting together we get: y(h) + y(p) => Acos2x + Bsin2x - (3x/4)cos2x :)>
How do you do the derivative of x^3 sin^2 (5x)?
The answer is 10x^3 sin5x cos5x + 3x^2 sin^2 (5x), but i just don't know how to get to that using the Chain Rule. I think I am messing up on one of the steps of the rule