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Question 1) (12 points) Consider the differential equation $x^2y'' - 3xy' + 3y = x^2$ on the domain $(0, \infty)$ (a) State the order of this differential equation. Then decide if this differential equation is linear or nonlinear. (b) Show that the family of curves $y = c_1x + c_2x^3 - x^2$ are solutions to this differential equation. (c) Prove using an appropriate Existence and Uniqueness Theorem that the family of curves in part (b) is actually the general solution to this differential equation. Now, for the final part of this problem, consider the specific initial value problem associated with this differential equation with the following initial conditions: $y(1) = 2$ $y'(1) = -1$ (d) Determine the unique solution to this IVP using the general solution from parts (b)/(c).

          Question 1) (12 points)
Consider the differential equation $x^2y'' - 3xy' + 3y = x^2$ on the domain $(0, \infty)$
(a) State the order of this differential equation. Then decide if this differential equation is
linear or nonlinear.
(b) Show that the family of curves $y = c_1x + c_2x^3 - x^2$ are solutions to this differential equation.
(c) Prove using an appropriate Existence and Uniqueness Theorem that the family of curves in
part (b) is actually the general solution to this differential equation.
Now, for the final part of this problem, consider the specific initial value problem associated with
this differential equation with the following initial conditions:
$y(1) = 2$
$y'(1) = -1$
(d) Determine the unique solution to this IVP using the general solution from parts (b)/(c).

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question 1 12 points consider the differential equation x2y 3xy 3y x2 on the domain 0 infty a state the order of this differential equation then decide if this differential equation is linea 79381

Added by Remedios J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Mary Wakumoto

05/24/2022

Step 1

In this case, the highest order derivative is $y''$, which is the second derivative. Therefore, the order of the differential equation is 2. To determine if the differential equation is linear or nonlinear, we check if the equation can be written in the Show more…

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Question 1) (12 points) Consider the differential equation $x^2y'' - 3xy' + 3y = x^2$ on the domain $(0, \infty)$ (a) State the order of this differential equation. Then decide if this differential equation is linear or nonlinear. (b) Show that the family of curves $y = c_1x + c_2x^3 - x^2$ are solutions to this differential equation. (c) Prove using an appropriate Existence and Uniqueness Theorem that the family of curves in part (b) is actually the general solution to this differential equation. Now, for the final part of this problem, consider the specific initial value problem associated with this differential equation with the following initial conditions: $y(1) = 2$ $y'(1) = -1$ (d) Determine the unique solution to this IVP using the general solution from parts (b)/(c).

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00:02 All right, so here we're taking a look at this differential equation.

00:06 And we noticed right away that instead of having just simple y, it's y squared.

00:12 This is what is making this differential equation nonlinear.

00:16 If we're going to be linear than all the derivatives are just simple, they're not square, they're not part of composites, they're just simple with powers of one only.

00:26 So we have a nonlinear differential equation.

00:29 The order is determined by the derivative, of the highest order.

00:33 We only have a first derivative.

00:35 So this is a first order differential equation.

00:41 All right.

00:43 So that was all part a, just kind of getting used to what is the kind of classification of this differential equation.

00:51 And part b is we're basically going to show that 1 over c plus x squared y equals.

01:02 Let me rewrite that with the y equals.

01:05 We want to show that this is a solution to our difference equation.

01:11 And so we're just going to basically find the derivative sub in and see if it works.

01:16 So we know why is this? i'm going to rewrite it as c plus x squared to the minus one.

01:21 So i can do some good old chain rule.

01:24 So do y -d -x then...

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