Separation of Variables

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 2.2 – Separation of Variables

Expected Educational Results

• Objective 5–1: I understand when the method of separation of variables is appropriate to solve ordinary differential equations.

• Objective 5–2: I can solve ordinary differential equations using the method of separation of variables.

• Objective 5–3: I can solve initial value problems using the method of separation of variables.

Calculus I

Derivatives

Differentiation Rules

General Formulas

• ${\displaystyle \frac{d}{dx}c=0}$$\displaystyle \frac{d}{dx}c=0$, where $c$$c$ is any constant.

• ${\displaystyle \frac{d}{dx}[cf(x)]=c{f}^{\prime }(x)}$$\displaystyle \frac{d}{dx}[cf(x)]=cf^{\,\prime}(x)$, where $c$$c$ is any constant.

• ${\displaystyle \frac{d}{dx}[f(x)\pm g(x)]=f^{\prime }(x)\pm g^{\prime }(x)}$$\displaystyle \frac{d}{dx}[f(x)\pm g(x)]=f^{\,\prime}(x)\pm g^{\,\prime}(x)$

• ${\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+g^{\prime }(x)f(x)}$$\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\,\prime}(x)g(x)+g^{\,\prime}(x)f(x)$

• ${\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f^{\prime }(x)g(x)-g^{\prime }(x)f(x)}{[g(x){]}^2}}$$\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f^{\,\prime}(x)g(x)-g^{\,\prime}(x)f(x)}{[g(x)]^2}$

• ${\displaystyle \frac{d}{dx}f(g(x))=f^{\prime }(g(x))g^{\prime }(x)}$$\displaystyle \frac{d}{dx}f(g(x))=f^{\,\prime}(g(x))g^{\,\prime}(x)$

• ${\displaystyle \frac{d}{dx}{x}^n=n{x}^{n-1}}$$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$\index{differentiation rules!power rule}

Exponential and Logarithmic Formulas

• ${\displaystyle \frac{d}{dx}{e}^x=e^x}$$\displaystyle \frac{d}{dx}e^x=e^x$

• ${\displaystyle \frac{d}{dx}{a}^x=\ln (a)a^x}$$\displaystyle \frac{d}{dx}a^x=\ln{(a)}a^x$, $a>0$$a>0$, $a\ne 1$$a\ne 1$

• ${\displaystyle \frac{d}{dx}\ln (x)=\frac{1}{x}}$$\displaystyle \frac{d}{dx}\ln{(x)}=\frac{1}{x}$

• ${\displaystyle \frac{d}{dx}{\log }_b(x)=\frac{1}{x\ln (b)}}$$\displaystyle \frac{d}{dx}\log_b{(x)}=\frac{1}{x\ln{(b)}}$, $b>0$$b>0$, $b\ne 1$$b\ne 1$

Trigonometric Formulas

• ${\displaystyle \frac{d}{dx}\sin (x)=\cos (x)}$$\displaystyle \frac{d}{dx}\sin{(x)}=\cos{(x)}$

• ${\displaystyle \frac{d}{dx}\cos (x)=-\sin (x)}$$\displaystyle \frac{d}{dx}\cos{(x)}=-\sin{(x)}$

• ${\displaystyle \frac{d}{dx}\tan (x)={\sec }^2(x)}$$\displaystyle \frac{d}{dx}\tan{(x)}=\sec^2{(x)}$

• ${\displaystyle \frac{d}{dx}\cot (x)=-{\csc }^2(x)}$$\displaystyle \frac{d}{dx}\cot{(x)}=-\csc^2{(x)}$

• ${\displaystyle \frac{d}{dx}\sec (x)=\sec (x)\tan (x)}$$\displaystyle \frac{d}{dx}\sec{(x)}=\sec{(x)}\tan{(x)}$

• ${\displaystyle \frac{d}{dx}\csc (x)=-\csc (x)\cot (x)}$$\displaystyle \frac{d}{dx}\csc{(x)}=-\csc{(x)}\cot{(x)}$

Inverse Trigonometric Formulas

• ${\displaystyle \frac{d}{dx}\arcsin (x)=\frac{1}{\sqrt{1-x^2}}}$$\displaystyle \frac{d}{dx}\arcsin{(x)}=\frac{1}{\sqrt{1-x^2}}$

• ${\displaystyle \frac{d}{dx}\arccos (x)=-\frac{1}{\sqrt{1-x^2}}}$$\displaystyle \frac{d}{dx}\arccos{(x)}=-\frac{1}{\sqrt{1-x^2}}$

• ${\displaystyle \frac{d}{dx}\arctan (x)=\frac{1}{1+x^2}}$$\displaystyle \frac{d}{dx}\arctan{(x)}=\frac{1}{1+x^2}$

• ${\displaystyle \frac{d}{dx}\text{arccot}(x)=-\frac{1}{1+x^2}}$$\displaystyle \frac{d}{dx}\text{arccot}{(x)}=-\frac{1}{1+x^2}$

• ${\displaystyle \frac{d}{dx}\text{arcsec}(x)=\frac{1}{|x|\sqrt{x^2-1}}}$$\displaystyle \frac{d}{dx}\text{arcsec}{(x)}=\frac{1}{|x|\sqrt{x^2-1}}$

• ${\displaystyle \frac{d}{dx}\text{arccsc}(x)=-\frac{1}{|x|\sqrt{x^2-1}}}$$\displaystyle \frac{d}{dx}\text{arccsc}{(x)}=-\frac{1}{|x|\sqrt{x^2-1}}$

Hyperbolic Formulas

• ${\displaystyle \frac{d}{dx}\sinh (x)=\cosh (x)}$$\displaystyle \frac{d}{dx}\sinh{(x)}=\cosh{(x)}$

• ${\displaystyle \frac{d}{dx}\cosh (x)=\sinh (x)}$$\displaystyle \frac{d}{dx}\cosh{(x)}=\sinh{(x)}$

• ${\displaystyle \frac{d}{dx}\tanh (x)={\text{sech}}^2(x)}$$\displaystyle \frac{d}{dx}\tanh{(x)}=\text{sech}^2{(x)}$

• ${\displaystyle \frac{d}{dx}\coth (x)=-{\text{csch}}^2(x)}$$\displaystyle \frac{d}{dx}\coth{(x)}=-\text{csch}^2{(x)}$

• ${\displaystyle \frac{d}{dx}\text{sech}(x)=-\text{sech}(x)\tanh (x)}$$\displaystyle \frac{d}{dx}\text{sech}{(x)}=-\text{sech}{(x)}\tanh{(x)}$

• ${\displaystyle \frac{d}{dx}\text{csch}(x)=-\text{csch}(x)\coth (x)}$$\displaystyle \frac{d}{dx}\text{csch}{(x)}=-\text{csch}{(x)}\coth{(x)}$

Inverse Hyperbolic Formulas

• ${\displaystyle \frac{d}{dx}\text{arcsinh}(x)=\frac{1}{\sqrt{1+x^2}}}$$\displaystyle \frac{d}{dx}\text{arcsinh}{(x)}=\frac{1}{\sqrt{1+x^2}}$

• ${\displaystyle \frac{d}{dx}\text{arccosh}(x)=\frac{1}{\sqrt{x^2-1}}}$$\displaystyle \frac{d}{dx}\text{arccosh}{(x)}=\frac{1}{\sqrt{x^2-1}}$

• ${\displaystyle \frac{d}{dx}\text{arctanh}(x)=\frac{1}{1-x^2}}$$\displaystyle \frac{d}{dx}\text{arctanh}{(x)}=\frac{1}{1-x^2}$

• ${\displaystyle \frac{d}{dx}\text{arccoth}(x)=\frac{1}{1-x^2}}$$\displaystyle \frac{d}{dx}\text{arccoth}{(x)}=\frac{1}{1-x^2}$

• ${\displaystyle \frac{d}{dx}\text{arcsech}(x)=-\frac{1}{x\sqrt{1-x^2}}}$$\displaystyle \frac{d}{dx}\text{arcsech}{(x)}=-\frac{1}{x\sqrt{1-x^2}}$

• ${\displaystyle \frac{d}{dx}\text{arccsch}(x)=-\frac{1}{|x|\sqrt{1+x^2}}}$$\displaystyle \frac{d}{dx}\text{arccsch}{(x)}=-\frac{1}{|x|\sqrt{1+x^2}}$

Antiderivatives

Antidifferentiation Rules

Basic Forms

• ${\displaystyle \int u^{n}du=\frac{u^{n+1}}{n+1}+C}$$\displaystyle \int{u^n}\,du=\frac{u^{n+1}}{n+1}+C$, $n\ne 1$$n\ne 1$

• ${\displaystyle \int u^{-1}du=\int \frac{1}{u}du=\ln |u|+C}$$\displaystyle \int{u^{-1}}\,du=\int{\frac{1}{u}}\,du=\ln{|u|}+C$

• ${\displaystyle \int a^{u}du=\frac{a^u}{\ln (a)}+C}$$\displaystyle \int{a^u}\,du=\frac{a^u}{\ln{(a)}}+C$, $a>0$$a>0$, $a\ne 1$$a\ne 1$

• ${\displaystyle \int \tan (u)du=\ln |\sec (u)|+C}$$\displaystyle \int{\tan{(u)}}\,du=\ln{|\sec{(u)}|}+C$

• ${\displaystyle \int \sec (u)du=\ln |\sec (u)+\tan (u)|+C}$$\displaystyle \int{\sec{(u)}}\,du=\ln{|\sec{(u)}+\tan{(u)}|}+C$

• ${\displaystyle \int \cot (u)du=\ln |\sin (u)|+C}$$\displaystyle \int{\cot{(u)}}\,du=\ln{|\sin{(u)}|}+C$

• ${\displaystyle \int \csc (u)du=\ln |\csc (u)-\cot (u)|+C}$$\displaystyle \int{\csc{(u)}}\,du=\ln{|\csc{(u)}-\cot{(u)}|}+C$

• ${\displaystyle \int \frac{1}{\sqrt{a^2-u^2}}du={\sin }^{-1}\left(\frac{u}{a}\right)+C}$$\displaystyle \int{\frac{1}{\sqrt{a^2-u^2}}}\,du=\sin^{-1}{\left(\frac{u}{a}\right)}+C$

• ${\displaystyle \int \frac{1}{a^2+u^2}du=\frac{1}{a}{\tan }^{-1}\left(\frac{u}{a}\right)+C}$$\displaystyle \int{\frac{1}{a^2+u^2}}\,du=\frac{1}{a}\tan^{-1}{\left(\frac{u}{a}\right)}+C$

Integration Strategy

This is a suggested strategy to use while attempting to evaluating an integral. This can be used for indefinite and definite integrals.

Fundamental Theorem of Calculus - Part I

Use known formulas from your knowledge of derivatives to find antiderivatives.

Example 01: ${\displaystyle \int \sec (x)\tan (x)dx}$$\displaystyle \int{\sec{(x)}\tan{(x)}\,dx}$

Since we know ${\displaystyle \frac{d}{dx}}\sec (x)=\sec (x)\tan (x)$$\dfrac{d}{dx}\sec{(x)}=\sec{(x)}\tan{(x)}$, then

$\Rightarrow \int {\displaystyle \frac{d}{dx}}\sec (x)dx=\int \sec (x)\tan (x)dx$$\Rightarrow\int{\dfrac{d}{dx}\sec{(x)}\,dx}=\int{\sec{(x)}\tan{(x)}\,dx}$, and

$\sec (x)+C=\int \sec (x)\tan (x)dx$$\sec{(x)}+C=\int{\sec{(x)}\tan{(x)}\,dx}$

Thus, ${\displaystyle \int \sec (x)\tan (x)dx=\sec (x)+C}$$\displaystyle \int{\sec{(x)}\tan{(x)}\,dx}=\sec{(x)}+C$

Example 02: ${\displaystyle \int e^{3x}dx}$$\displaystyle \int{e^{3x}\,dx}$

Since we know ${\displaystyle \frac{d}{dx}}{e}^{3x}=3e^{3x}$$\dfrac{d}{dx}e^{3x}=3e^{3x}$, then

$\Rightarrow \int {\displaystyle \frac{d}{dx}}{e}^{3x}dx=\int 3e^{3x}dx$$\Rightarrow\int{\dfrac{d}{dx}e^{3x}\,dx}=\int{3e^{3x}\,dx}$, which is equivalent to

$\Rightarrow \int \frac{d}{dx}{e}^{3x}dx=3\int e^{3x}dx$$\Rightarrow\int{\frac{d}{dx}e^{3x}\,dx}=3\int{e^{3x}\,dx}$, and

$\Rightarrow {\displaystyle \frac{1}{3}}\int \frac{d}{dx}{e}^{3x}dx=\int e^{3x}dx$$\Rightarrow\dfrac{1}{3}\int{\frac{d}{dx}e^{3x}\,dx}=\int{e^{3x}\,dx}$

Thus, ${\displaystyle \int e^{3x}dx=\frac{1}{3}{e}^{3x}+C}$$\displaystyle \int{e^{3x}\,dx}=\frac{1}{3}e^{3x}+C$

Rewrite the Integrand

Use algebra or trigonometry to rewrite the integrand.

Example 03: ${\displaystyle \int \frac{x-1}{x^2-1}dx}$$\displaystyle \int{\frac{x-1}{x^2-1}\,dx}$

$\int {\displaystyle \frac{x-1}{x^2-1}}dx=\int {\displaystyle \frac{x-1}{(x-1)(x+1)}}dx$$\int{\dfrac{x-1}{x^2-1}\,dx}=\int{\dfrac{x-1}{(x-1)(x+1)}\,dx}$

$\Rightarrow \int {\displaystyle \frac{1}{x+1}}dx$$\Rightarrow\int{\dfrac{1}{x+1}\,dx}$ $\Rightarrow \ln (x+1)+C$$\Rightarrow\ln{(x+1)}+C$

Example 04: ${\displaystyle \int \frac{x^2}{x^2-1}dx}$$\displaystyle \int{\frac{x^2}{x^2-1}\,dx}$

$\int {\displaystyle \frac{x^2}{x^2-1}}dxd=\int {\displaystyle \frac{x^2+0}{x^2-1}}dx$$\int{\dfrac{x^2}{x^2-1}\,dx}d=\int{\dfrac{x^2{\color{gsured}+0}}{x^2-1}\,dx}$

$\Rightarrow \int {\displaystyle \frac{x^2-1+1}{x^2-1}}dx$$\Rightarrow\int{\dfrac{x^2{\color{gsured}-1+1}}{x^2-1}\,dx}$

$\Rightarrow \int ({\displaystyle \frac{x^2-1}{x^2-1}}+{\displaystyle \frac{1}{x^2-1}})dx$$\Rightarrow\int{\left(\dfrac{x^2-1}{x^2-1}+\dfrac{1}{x^2-1}\right)\,dx}$

$\Rightarrow \int (1+{\displaystyle \frac{1}{x^2-1}})dx$$\Rightarrow\int{\left(1+\dfrac{1}{x^2-1}\right)\,dx}$

$\Rightarrow \int \left(1-{\displaystyle \frac{1}{1-x^2}}\right)dx$$\Rightarrow\int{\left(1{\color{gsured}{-}}\dfrac{1}{{\color{gsured}1-x^2}}\right)\,dx}$

$\Rightarrow x-\text{arctanh}(x)+C$$\Rightarrow x-\text{arctanh}(x)+C$

Example 05: ${\displaystyle \int \frac{\tan (x)}{\sin (x)}dx}$$\displaystyle \int{\frac{\tan{(x)}}{\sin{(x)}}\,dx}$

$\int {\displaystyle \frac{\tan (x)}{\sin (x)}}dx=\int {\displaystyle \frac{\sin (x)}{\cos (x)}}{\displaystyle \frac{1}{\sin (x)}}dx$$\int{\dfrac{\tan{(x)}}{\sin{(x)}}\,dx}=\int{\dfrac{\sin{(x)}}{\cos{(x)}}\dfrac{1}{\sin{(x)}}\,dx}$

$\Rightarrow \int {\displaystyle \frac{1}{\cos (x)}}dx$$\Rightarrow\int{\dfrac{1}{\cos{(x)}}\,dx}$

$\Rightarrow \int \sec (x)dx$$\Rightarrow\int{\sec{(x)}\,dx}$

$\Rightarrow \ln |\sec (x)+\tan (x)|+C$$\Rightarrow\ln{\left|\sec{(x)}+\tan{(x)}\right|}+C$

Example 06: ${\displaystyle \int {\sin }^2(x)dx}$$\displaystyle \int{\sin^2{(x)}\,dx}$

$\int {\sin }^2(x)dx=\int {\displaystyle \frac{1}{2}}(1-\cos (2x))dx$$\int{\sin^2{(x)}\,dx}=\int{\dfrac{1}{2}\left(1-\cos{(2x)}\right)\,dx}$

$\Rightarrow {\displaystyle \frac{1}{2}}(x-{\displaystyle \frac{1}{2}}\sin (2x))+C$$\Rightarrow\dfrac{1}{2}\left(x-\dfrac{1}{2}\sin{(2x)}\right)+C$

Additional Methods

• u Substitution: http://www.jjw3.com/r/uSub.pdf

• Integration by Parts: http://www.jjw3.com/r/byParts.pdf

• Trigonometric Substitution: http://www.jjw3.com/r/trigSub.pdf

• Method of Partial Fractions: http://www.jjw3.com/r/partialFractions.pdf

• Use multiple methods

• Use a Table of Integrals or CAS

Practice 01

Evaluate the following integrals:

1. ${\displaystyle \int (3x^2+x-5)e^{-x}dx}$$\displaystyle \int{(3x^2+x-5)e^{-x}\,dx}$

2. ${\displaystyle \int \frac{1}{1-x^2}dx}$$\displaystyle \int{\frac{1}{1-x^2}\,dx}$

3. ${\displaystyle \int \frac{x}{1-x^2}dx}$$\displaystyle \int{\frac{x}{1-x^2}\,dx}$

4. ${\displaystyle \int \frac{x^2}{1-x^2}dx}$$\displaystyle \int{\frac{x^2}{1-x^2}\,dx}$

5. ${\displaystyle \int \frac{x^3}{1-x^2}dx}$$\displaystyle \int{\frac{x^3}{1-x^2}\,dx}$

6. ${\displaystyle \int \frac{x^2}{1+x^2}dx}$$\displaystyle \int{\frac{x^2}{1+x^2}\,dx}$

7. ${\displaystyle \int \frac{x^3}{1+x^3}dx}$$\displaystyle \int{\frac{x^3}{1+x^3}\,dx}$

8. ${\displaystyle \int \frac{1+x^2}{1+x}dx}$$\displaystyle \int{\frac{1+x^2}{1+x}\,dx}$

9. ${\displaystyle \int \frac{{\sec }^2(x)}{1+\tan (x)}dx}$$\displaystyle \int{\frac{\sec^2{(x)}}{1+\tan{(x)}}\,dx}$

10. ${\displaystyle \int \frac{1-\sin (x)}{\cos (x)}dx}$$\displaystyle \int{\frac{1-\sin{(x)}}{\cos{(x)}}\,dx}$

11. ${\displaystyle \int \frac{x}{x^2-3x+2}dx}$$\displaystyle \int{\frac{x}{x^2-3x+2}\,dx}$

12. ${\displaystyle \int \frac{x^2-3x+2}{x}dx}$$\displaystyle \int{\frac{x^2-3x+2}{x}\,dx}$

13. ${\displaystyle \int {\sin }^{-1}(x)dx}$$\displaystyle \int{\sin^{-1}{(x)}\,dx}$

14. ${\displaystyle \int {\sin }^3(x){\cos }^5(x)dx}$$\displaystyle \int{\sin^3{(x)}\cos^5{(x)}\,dx}$

15. ${\displaystyle \int {\sin }^3(x){\cos }^2(x)dx}$$\displaystyle \int{\sin^3{(x)}\cos^2{(x)}\,dx}$

16. ${\displaystyle \int {\tan }^3(x){\sec }^3(x)dx}$$\displaystyle \int{\tan^3{(x)}\sec^3{(x)}\,dx}$

17. ${\displaystyle \int {\tan }^4(x){\sec }^4(x)dx}$$\displaystyle \int{\tan^4{(x)}\sec^4{(x)}\,dx}$

18. ${\displaystyle \int {\cos }^4(x)dx}$$\displaystyle \int{\cos^4{(x)}\,dx}$

19. ${\displaystyle \int \frac{1}{x\sqrt{\ln (x)}}dx}$$\displaystyle \int{\frac{1}{x\sqrt{\ln{(x)}}}\,dx}$

20. ${\displaystyle \int x\ln (x)dx}$$\displaystyle \int{x\ln{(x)}\,dx}$

21. ${\displaystyle \int e^{\sqrt{x}}dx}$$\displaystyle \int{e^{\sqrt{x}}\,dx}$

22. ${\displaystyle \int \sqrt{x^2-4x}dx}$$\displaystyle \int{\sqrt{x^2-4x}\,dx}$

23. ${\displaystyle \int \frac{1}{x^2\sqrt{x^2+1}}dx}$$\displaystyle \int{\frac{1}{x^2\sqrt{x^2+1}}\,dx}$

24. ${\displaystyle \int e^{3x}\cos (5x)dx}$$\displaystyle \int{e^{3x}\cos{(5x)}\,dx}$

25. ${\displaystyle \int \frac{1}{e^{-x}+e^x}dx}$$\displaystyle \int{\frac{1}{e^{-x}+e^x}\,dx}$

26. ${\displaystyle \int \frac{x+1}{x^2+4}dx}$$\displaystyle \int{\frac{x+1}{x^2+4}\,dx}$

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Last Modified: Monday, 31 August 2020 12:24 EDT