/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Solve each of the following by t... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 23

Solve each of the following by two methods (see Exercise \(21(b))\) : (a) \(\left(x^{2}+2 y^{2}\right) d x+\left(4 x y-y^{2}\right) d y=0\) (b) \(\left(2 x^{2}+2 x y+y^{2}\right) d x+\left(x^{2}+2 x y\right) d y=0\)

Short Answer

Expert verified
For equation (a): \[\left(x^{2}+2 y^{2}\right) d x+\left(4 x y-y^{2}\right) d y=0\] The solution is: \[x^2 + 2y^2 + 2x(y^2) - \frac{y^3}{3} = C\] For equation (b): \[\left(2 x^{2}+2 x y+y^{2}\right) d x+\left(x^{2}+2 x y\right) d y=0\] The solution is: \[2x^2 + 2xy + y^2 + x^2y + x(y^2) = C\]

Step by step solution

01

Test for exactness

Check if \(M = x^2 + 2y^2\) and \(N = 4xy - y^2\) are exact. Compute the partial derivatives \(\frac{\partial M}{\partial y} = 4y\) \(\frac{\partial N}{\partial x} = 4y\) Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the given equation is an exact differential equation.
02

Exact differential equation method

Rewrite the equation as: \(d(x^2 + 2y^2) + (4xy - y^2) dy = 0\) Integrate both sides with respect to x: \(\int [d(x^2 + 2y^2) + (4xy - y^2) dy] \,dx\) \(x^2 + 2y^2 + \int (4xy - y^2) dy = C\) Now, integrate \(4xy - y^2\) with respect to y: \(x^2 + 2y^2 + 2x(y^2) - \frac{y^3}{3} = C\)
03

Integrating factor method

Since the equation is exact, there is no need to use the integrating factor method. The solution is: \(x^2 + 2y^2 + 2x(y^2) - \frac{y^3}{3} = C\) #### Part (b) ####
04

Test for exactness

Check if \(M = 2x^2 + 2xy + y^2\) and \(N = x^2 + 2xy\) are exact. Compute the partial derivatives \(\frac{\partial M}{\partial y} = 2x + 2y\) \(\frac{\partial N}{\partial x} = 2x + 2y\) Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the given equation is an exact differential equation.
05

Exact differential equation method

Rewrite the equation as: \(d(2x^2 + 2xy + y^2) + (x^2 + 2xy) dy = 0\) Integrate both sides with respect to x: \(\int [d(2x^2 + 2xy + y^2) + (x^2 + 2xy) dy] \,dx\) \(2x^2 + 2xy + y^2 + \int (x^2 + 2xy) dy = C\) Now, integrate \(x^2 + 2xy\) with respect to y: \(2x^2 + 2xy + y^2 + x^2y + x(y^2) = C\)
06

Integrating factor method

Since the equation is exact, there is no need to use the integrating factor method. The solution is: \(2x^2 + 2xy + y^2 + x^2y + x(y^2) = C\)

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Most popular questions from this chapter

Solve the given differential equations. \(\frac{d y}{d x}+3 y=3 x^{2} e^{-3 x}\)

(a) Prove that if \(f\) and \(g\) are two different solutions of $$ \frac{d y}{d x}+P(x) y=Q(x) $$ then \(f-g\) is a solution of the equation $$ \frac{d y}{d x}+P(x) y=0 $$ (b) Thus show that if \(f\) and \(g\) are two different solutions of Equation (A) and \(c\) is an arbitrary constant, then $$ c(f-g)+f $$ is a one-parameter family of solutions of (A).

Solve each of the differential equations. \((x+4)\left(y^{2}+1\right) d x+y\left(x^{2}+3 x+2\right) d y=0\)

Solve each differential equation by first finding an integrating factor. \(\left(5 x y+4 y^{2}+1\right) d x+\left(x^{2}+2 x y\right) d y=0\)

(a) Let \(f_{1}\) be a solution of $$ \frac{d y}{d x}+P(x) y=Q_{1}(x) $$ and \(f_{2}\) be a solution of $$ \frac{d y}{d x}+P(x) y=Q_{2}(x) $$ where \(P, Q_{1}\), and \(Q_{2}\) are all defined on the same real interval 1 . Prove that \(f_{1}+f_{2}\) is a solution of $$ \frac{d y}{d x}+P(x) y=Q_{1}(x)+Q_{2}(x) $$ on \(I .\) (b) Use the result of (a) to solve the equation $$ \frac{d y}{d x}+y=2 \sin x+5 \sin 2 x $$
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