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karush
Gold Member
MHB
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OK going to comtinue with these till I have more confidence with it
$$\dfrac{dy}{dx}=2 (1+x) (1+y^2), \qquad y(0)=0$$
separate
$$(1+y^2)\, dy=(2+2x)\, dx$$
Last edited:
$(1+x)^2 $ |
You aren't lost. You got the correct answer!karush said:so...then,,,,
$$\arctan \left(y\right)=x^2+2x$$
then
$$y=\tan(x^2+2x)$$
there is no book answer to this
ok sorry I am kinda lost
MarkFL said:The ODE associated with this IVP is separable. I would next write:
\(\displaystyle \int_0^y \frac{1}{u^2+1}\,du=2\int_0^x v+1\,dv\)
And...GO!
A first order IVP, or initial value problem, is a type of differential equation that involves finding a function that satisfies both a differential equation and an initial condition. The initial condition provides a starting point for the solution of the differential equation.
To solve a first order IVP, you can use various methods such as separation of variables, integrating factors, or the method of undetermined coefficients. Each method involves manipulating the differential equation to isolate the dependent and independent variables and then integrating to find the solution.
The minimum of a solution to a first order IVP is the smallest value that the solution takes on over a given interval. This can be found by taking the derivative of the solution and setting it equal to zero, then solving for the value of the independent variable at which the minimum occurs.
To determine where the minimum of a solution to a first order IVP occurs, you can take the derivative of the solution and set it equal to zero. Then, solve for the value of the independent variable at which the derivative is equal to zero. This value will correspond to the location of the minimum.
Yes, the minimum of a solution to a first order IVP can occur at multiple points if the solution has multiple local minima. In this case, the derivative of the solution will be equal to zero at each of these points, and the minimum will occur at each of these points.