Show that y= A sin 2x + 3 cos 2x is solution of y" + 4y= 0
We have a second order differential equation and we have been given the general solution. Our job is to show that the solution is correct.We will do this substituting the answer into the original 2nd order differential equation. We need to find the second derivative of y y= A sin 2x + 3 cos 2x ..........................(1) First Derivative= y' = 2* (A cos2x) - 2 * ( 3sin2x)= 2A cos2x - 6 sin2x Second derivative= y"= - 2* (2A sin2x) - 2*(6cos2x)= -4A sin2x - 12cos2x ...........(2) Now substitute back in y"+ 4y. To be the solution it should come to zero. LHS= y"+ 4y= -4A sin2x - 12cos2x + 4*(A sin 2x + 3 cos 2x) ...............From (1) & (2) =-4A sin2x - 12cos2x + 4A sin2x+12cos2x = 0 Hence y= A sin 2x + 3 cos 2x is solution of y" + 4y= 0
Differentiate ln ( sinx -cotx)
Let f(x)=ln ( sinx -cotx) then derivative of f(x) will be, f'(x) = (1/sinx-cotx) *(cosx + csc^2 x) .... (1) For differentiating logarithm , differentiating the logarithm (i.e. the outside function) you need to substitute the inside function into the derivative and then multiply by the derivative of inside function Here sinx - cotx is inside function so the answer is (1)
Find the quadratic equation with real coefficients which has 3 - 2i as a root (i = √-1).
According to the problem, coefficients of the required quadratic equation are real and its one root is 3 - 2i. Hence, the other root of the required equation is 3 - 2i (Since, the complex roots always occur in pairs, so other root is 3 + 2i. Now, the sum of the roots of the required equation = 3 - 2i + 3 + 2i = 6 And, product of the roots = (3 + 2i)(3 - 2i) = 322 - (2i)22 = 9 - 4i22 = 9 -4(-1) = 9 + 4 = 13 Hence, the equation is x22 - (Sum of the roots)x + product of the roots = 0 i.e., x22 - 6x + 13 = 0 Therefore, the required equation is x22 - 6x + 13 = 0.