Can someone help me with this question? I seem to be getting a negative answer every time :(
2 answers:
You get a negative value for x² because there are no real-number solutions to the set of equations. (The graphs do not intersect.)
Solve for x^2:
... x^2 = (y^2 -52)/3 = (16-52)/3) = -12
... x = 2√(-3) = ±i·2√3
Complex numbers cannot be compared to zero. (x > 0 has no meaning in this context.) It seems possible you want the complex solution with the positive imaginary part: x = i·2√3
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Answer:
Step-by-step explanation:
(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.
(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.
(C) Example of a second order linear ODE:
M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)
The equation will be homogeneous if K(t)=0 and heterogeneous if
Example of a second order nonlinear ODE:
(D) Example of a nonlinear fourth order ODE: