ITEMSINFOSolve the initial value problem yy'+¤ = /x² + y² with y(2) = v21.a. To solve this, we should use the substitutionhelp (formulas)U =u' =help (formulas)Enter derivatives using prime notation (e.g.. you would enter y' for ).b. After the substitution from the previous part, we obtain the following linear differential equation in x, u, u'.help (equations)c. The solution to the original initial value problem is described by the following equation in r, y.help (equations)Submit answerAnswers (in progress)Answor

given equation is yy'+x=x2+y2 at x=2,…

Answered: Solve the initial value problem yy' + ¤…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

ITEMS
INFO
Solve the initial value problem yy'+¤ = /x² + y² with y(2) = v21.
a. To solve this, we should use the substitution
help (formulas)
U =
u' =
help (formulas)
Enter derivatives using prime notation (e.g.. you would enter y' for ).
b. After the substitution from the previous part, we obtain the following linear differential equation in x, u, u'.
help (equations)
c. The solution to the original initial value problem is described by the following equation in r, y.
help (equations)
Submit answer
Answers (in progress)
Answor

Expert Solution

Step 1

given equation is

$y y' + x = \sqrt{x^{2} + y^{2}}$

at $x = 2, y = \sqrt{21}$

Let

$u = \sqrt{x^{2} + y^{2}} u^{2} = x^{2} + y^{2}$

differentiating with respect to x

$\begin{array}{rcl} u^{2} & = & x^{2} + y^{2} \\ 2 u u' & = & 2 x + 2 y y' \\ u u' & = & x + y y' \\ u' & = & \frac{x + y y'}{u} \\ u' & = & \frac{x + y y'}{\sqrt{x^{2} + y^{2}}} \end{array}$

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Differential Equation$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions