/*! 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 initial-value problems. $$ (3 x-y-6) d x+(x+y+2) d y=0, \quad y(2)=-2 $$

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

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

Solve the initial-value problems. The equation $$ \frac{d y}{d x}=A(x) y^{2}+B(x) y+C(x) $$ is called Riccati's equation. (a) Show that if \(A(x)=0\) for all \(x\), then Equation (A) is a linear equation, whereas if \(C(x)=0\) for all \(x\), then Equation (A) is a Bernoulli equation. (b) Show that if \(f\) is any solution of Equation (A), then the transformation $$ y=f+\frac{1}{v} $$ reduces (A) to a linear equation in \(v\). In each of Exercises \(39-41\), use the result of Exercise 38 (b) and the given solution to find a one-parameter family of solutions of the given Riccati

$$ \left(x^{2}+3 y^{2}\right) d x-2 x y d y=0, \quad y(2)=6 $$
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