ƏM5) For the non exact equation Mdx + Ndy = 0, if ÷(N - ) is a function of y alone, say fV)M \axthenis an integrating factor.6) The general solution of an ODE y(xy Sin(xy) + Cos(xy))dx + x(xy Sin(xy) – Cos(xy))dy =0, is7) If the equation + Px = Q, where P & Q are functions of y, then integrating factor =%3Ddy8)is the solution of an ODE x+ y = y² log(x).9) The solution of an initial value problem y3 +y = x; y(1) = 2 is10) The equation x²p² + xyp – 6y² = 0, where p = " is solvable for

Answered: ƏM 5) For the non exact equation Mdx +…

ƏM 5) For the non exact equation Mdx + Ndy = 0, if –M - ) is a function of y alone, say fV) M \ax then is an integrating factor.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

Question

ƏM
5) For the non exact equation Mdx + Ndy = 0, if ÷(N - ) is a function of y alone, say fV)
M \ax
then
is an integrating factor.
6) The general solution of an ODE y(xy Sin(xy) + Cos(xy))dx + x(xy Sin(xy) – Cos(xy))dy =
0, is
7) If the equation + Px = Q, where P & Q are functions of y, then integrating factor =
%3D
dy
8)
is the solution of an ODE x+ y = y² log(x).
9) The solution of an initial value problem y3 +y = x; y(1) = 2 is
10) The equation x²p² + xyp – 6y² = 0, where p = " is solvable for

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