# 2.6E: Integrating Factors (Exercises)

- Page ID
- 18249

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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’s 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)?