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2.6E: Integrating Factors (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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[exer:2.6.1}

Verify that $$\mu(x,y)=y$$ is an integrating factor for

$y\,dx+\left(2x+{1\over y}\right)\,dy=0 \eqno{\rm (A)}$

on any open rectangle that does not intersect the $$x$$ axis or, equivalently, that

$y^2\,dx+(2xy+1)\,dy=0 \eqno{\rm (B)}$

is exact on any such rectangle.

Verify that $$y\equiv0$$ is a solution of (B), but not of (A).

Show that

$y(xy+1)=c \eqno{\rm (C)}$

is an implicit solution of (B), and explain why every differentiable function $$y=y(x)$$ other than $$y\equiv0$$ that satisfies (C) is also a solution of (A).

[exer:2.6.2}

Verify that $$\mu(x,y)=1/(x-y)^2$$ is an integrating factor for

$-y^2\,dx+x^2\,dy=0 \eqno{\rm (A)}$

on any open rectangle that does not intersect the line $$y=x$$ or, equivalently, that

$-{y^2\over(x-y)^2}\,dx+{x^2\over(x-y)^2}\,dy=0 \eqno{\rm (B)}$

is exact on any such rectangle.

Use Theorem [thmtype:2.2.1} to show that

${xy\over(x-y)}=c \eqno{\rm (C)}$

is an implicit solution of (B), and explain why it is also an implicit solution of (A)

Verify that $$y=x$$ is a solution of (A), even though it can’t be obtained from (C).

 [exer:2.6.3} $$y\,dx-x\,dy=0$$ [exer:2.6.4} $$3x^2y\,dx+2x^3\,dy=0$$
 [exer:2.6.5} $$2y^3\,dx+3y^2\,dy=0$$ [exer:2.6.6} $$(5xy+2y+5)\,dx+2x\,dy=0$$

[exer:2.6.7} $$(xy+x+2y+1)\,dx+(x+1)\,dy=0$$

[exer:2.6.8} $$(27xy^2+8y^3)\,dx+(18x^2y+12xy^2)\,dy=0$$

[exer:2.6.9} $$(6xy^2+2y)\,dx+(12x^2y+6x+3)\,dy=0$$

[exer:2.6.10} $$y^2\,dx+\dst{\left(xy^2+3xy+{1\over y}\right)\,dy=0}$$

[exer:2.6.11} $$(12x^3y+24x^2y^2)\,dx+(9x^4+32x^3y+4y)\,dy=0$$

[exer:2.6.12} $$(x^2y+4xy+2y)\,dx+(x^2+x)\,dy=0$$

[exer:2.6.13} $$-y\,dx+(x^4-x)\,dy=0$$

[exer:2.6.14} $$\cos x\cos y\,dx +(\sin x\cos y-\sin x\sin y+y)\,dy=0$$

[exer:2.6.15} $$(2xy+y^2)\,dx+(2xy+x^2-2x^2y^2-2xy^3)\,dy=0$$

[exer:2.6.16} $$y\sin y\,dx+x(\sin y-y\cos y)\,dy=0$$

[exer:2.6.17} $$y(1+5\ln|x|)\,dx+4x\ln|x|\,dy=0$$

[exer:2.6.18} $$(\alpha y+ \gamma xy)\,dx+(\beta x+ \delta xy)\,dy=0$$

[exer:2.6.19} $$(3x^2y^3-y^2+y)\,dx+(-xy+2x)\,dy=0$$

[exer:2.6.20} $$2y\,dx+ 3(x^2+x^2y^3)\,dy=0$$

[exer:2.6.21} $$(a\cos xy-y\sin xy)\,dx+(b\cos xy-x\sin xy)\, dy=0$$

[exer:2.6.22} $$x^4y^4\,dx+x^5y^3\,dy=0$$

[exer:2.6.23} $$y(x\cos x+2\sin x)\,dx+x(y+1)\sin x\,dy=0$$

[exer:2.6.24} $$(x^4y^3+y)\,dx+(x^5y^2-x)\,dy=0; \quad \{-1\le x\le1,-1\le y\le1\}$$

[exer:2.6.25} $$(3xy+2y^2+y)\,dx+(x^2+2xy+x+2y)\,dy=0; \quad \{-2\le x\le2,-2\le y\le2\}$$

[exer:2.6.26} $$(12 xy+6y^3)\,dx+(9x^2+10xy^2)\,dy=0; \quad \{-2\le x\le2,-2\le y\le2\}$$

[exer:2.6.27} $$(3x^2y^2+2y)\,dx+ 2x\,dy=0; \quad \{-4\le x\le4,-4\le y\le4\}$$

[exer:2.6.28} Suppose $$a$$, $$b$$, $$c$$, and $$d$$ are constants such that $$ad-bc\ne0$$, and let $$m$$ and $$n$$ be arbitrary real numbers. Show that

$(ax^my+by^{n+1})\,dx+(cx^{m+1}+dxy^n)\,dy=0$

has an integrating factor $$\mu(x,y)=x^\alpha y^\beta$$.

[exer:2.6.29} Suppose $$M$$, $$N$$, $$M_x$$, and $$N_y$$ are continuous for all $$(x,y)$$, and $$\mu=\mu(x,y)$$ is an integrating factor for

$M(x,y)\,dx+N(x,y)\,dy=0. \eqno{\rm (A)}$

Assume that $$\mu_x$$ and $$\mu_y$$ are continuous for all $$(x,y)$$, and suppose $$y=y(x)$$ is a differentiable function such that $$\mu(x,y(x))=0$$ and $$\mu_x(x,y(x))\ne0$$ for all $$x$$ in some interval $$I$$. Show that $$y$$ is a solution of (A) on $$I$$.

[exer:2.6.30} According to Theorem [thmtype:2.1.2}, the general solution of the linear nonhomogeneous equation

$y'+p(x)y=f(x) \eqno{\rm(A)}$

is

$y=y_1(x)\left(c+\int f(x)/y_1(x)\,dx\right), \eqno{\rm(B)}$

where $$y_1$$ is any nontrivial solution of the complementary equation $$y'+p(x)y=0$$. In this exercise we obtain this conclusion in a different way. You may find it instructive to apply the method suggested here to solve some of the exercises in Section 2.1.

Rewrite (A) as

$[p(x)y-f(x)]\,dx+\,dy=0, \eqno{\rm(C)}$

and show that $$\mu=\pm e^{\int p(x)\,dx}$$ is an integrating factor for (C).

Multiply (A) through by $$\mu=\pm e^{\int p(x)\,dx}$$ and verify that the resulting equation can be rewritten as

$(\mu(x)y)'=\mu(x)f(x).$

Then integrate both sides of this equation and solve for $$y$$ to show that the general solution of (A) is

$y={1\over\mu(x)}\left(c+\int f(x)\mu(x)\,dx\right).$

Why is this form of the general solution equivalent to (B)?