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Q 2.6-39E

Expert-verified
Found in: Page 77

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

Question: In Problems 33–40, solve the equation given in: Problem 7.

The solution of the equation given in problem 7 is $cos\left(x+y\right)+sin\left(x+y\right)=C{e}^{x-y}$.

See the step by step solution

Step 1: Concept and definition

When the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$ can be expressed as a function of the combination $ax+by$ , where a and b are constants, that is, $\frac{dy}{dx}=G\left(ax+by\right)$ , then the substitution $z=ax+by$ transforms the given equation into a separable equation.

Step2: Analyzing the given statement

One has to solve the equation given in problem 7, i.e.,

$\mathrm{cos}\left(x+y\right)dy=\mathrm{sin}\left(x+y\right)dx\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Step 3: Rewriting the equation (1) in the form  dydx=G(ax+by)

Rewriting the equation (1) as,

$\frac{dy}{dx}=\frac{\mathrm{sin}\left(x+y\right)}{\mathrm{cos}\left(x+y\right)}$

i.e., $\frac{dy}{dx}=\mathrm{tan}\left(x+y\right)\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Step 4: Substituting z = x + y  in equation (2) to transform the equation in the variables z and x

Taking $z=x+y$

Therefore, $dz=dx+dy$

Dividing both sides by dx,

$\begin{array}{c}\frac{dz}{dx}=1+\frac{dy}{dx}\\ \frac{dy}{dx}=\frac{dz}{dx}-1\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Substitute z = x + y in equation (2) and the value of dy/dx from equation (3),

Step 5: Find the solution of equation  dzdx=1+tanz

Separate the variables in equation (4),

$\begin{array}{c}\frac{dz}{dx}-1=\mathrm{tan}z\\ \frac{dz}{dx}=1+\mathrm{tan}z\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)\\ \frac{dz}{1+\mathrm{tan}z}=dx\end{array}$

Integrating both sides,

$\int \frac{dz}{1+\mathrm{tan}z}=\int dx\cdot \cdot \cdot \cdot \cdot \cdot \left(5\right)$

Put tan z = t

$\begin{array}{c}{\mathrm{sec}}^{2}zdz=dt\\ \left(1+{\mathrm{tan}}^{2}z\right)dz=dt\\ \left(1+{t}^{2}\right)dz=dt\\ dz=\frac{dt}{\left(1+{t}^{2}\right)}\end{array}$

Thus, equation (5) becomes,

$\int dx=\int \frac{dt}{\left(1+t\right)\left(1+{t}^{2}\right)}\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(6\right)$

The right hand side of the equation (6) can be solved by using partial fraction,

Therefore,

$\frac{1}{\left(1+t\right)\left(1+{t}^{2}\right)}=\frac{A}{1+t}+\frac{Bt+{C}_{1}}{1+{t}^{2}}$

Multiply both sides by $\left(1+t\right)\left(1+{t}^{2}\right)$

$1=A\left(1+{t}^{2}\right)+Bt\left(1+t\right)+{C}_{1}\left(1+t\right)\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(7\right)$

Put t = -1

$\begin{array}{c}1=2A+0+0\\ A=\frac{1}{2}\end{array}$

Equating coefficients in equation (7) of:

t2) $A+B=0\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \left(8\right)$

t ) $B+{C}_{1}=0\cdot \cdot \cdot \cdot \cdot \cdot \left(9\right)$

From equations (8) and (9),

$\begin{array}{c}\frac{1}{2}+B=0\\ B=-\frac{1}{2}\\ -\frac{1}{2}+{C}_{1}=0\\ {C}_{1}=\frac{1}{2}\end{array}$

Therefore, equation (6) becomes,

$\begin{array}{c}\int dx=\frac{1}{2}\int \frac{dt}{\left(1+t\right)}-\frac{1}{2}\int \frac{t}{1+{t}^{2}}dt+\frac{1}{2}\int \frac{dt}{1+{t}^{2}}\\ \int dx=\frac{1}{2}\int \frac{dt}{\left(1+t\right)}-\frac{1}{4}\int \frac{2t}{1+{t}^{2}}dt+\frac{1}{2}\int \frac{dt}{1+{t}^{2}}\\ x=\frac{1}{2}\mathrm{ln}\left|1+t\right|-\frac{1}{4}\mathrm{ln}\left|1+{t}^{2}\right|+\frac{1}{2}{\mathrm{tan}}^{-1}t+\mathrm{ln}{C}_{2}\end{array}$

Where, C2 is the constant of integration.

Put t = tan z

$\begin{array}{c}x=\frac{1}{2}\mathrm{ln}\left|1+\mathrm{tan}z\right|-\frac{1}{4}\mathrm{ln}\left|1+{\mathrm{tan}}^{2}z\right|+\frac{1}{2}{\mathrm{tan}}^{-1}\left(\mathrm{tan}z\right)+\mathrm{ln}{C}_{2}\\ x=\frac{1}{2}\mathrm{ln}\left|1+\mathrm{tan}z\right|-\frac{1}{4}\mathrm{ln}\left|1+{\mathrm{tan}}^{2}z\right|+\frac{1}{2}z+\mathrm{ln}{C}_{2}\\ 2x=\mathrm{ln}\left|1+\mathrm{tan}z\right|-\frac{1}{2}\mathrm{ln}\left|1+{\mathrm{tan}}^{2}z\right|+z+\mathrm{ln}{C}_{2}\\ 2x-z=\mathrm{ln}\left|1+\mathrm{tan}z\right|-\mathrm{ln}{\left(1+{\mathrm{tan}}^{2}z\right)}^{1/2}+\mathrm{ln}{C}_{2}\end{array}$

$\begin{array}{c}2x-z=\mathrm{ln}\frac{\left(1+\mathrm{tan}z\right)}{{\left(1+{\mathrm{tan}}^{2}z\right)}^{1/2}}+\mathrm{ln}{C}_{2}\\ 2x-z=\mathrm{ln}\frac{{C}_{2}\left(1+\mathrm{tan}z\right)}{{\left(1+{\mathrm{tan}}^{2}z\right)}^{1/2}}\end{array}$

Put z = x + y

$\begin{array}{c}2x-\left(x+y\right)=\mathrm{ln}\frac{{C}_{2}\left(1+\mathrm{tan}\left(x+y\right)\right)}{{\left(1+{\mathrm{tan}}^{2}\left(x+y\right)\right)}^{1/2}}\\ 2x-x-y=\mathrm{ln}\frac{{C}_{2}\left(1+\mathrm{tan}\left(x+y\right)\right)}{{\left(1+{\mathrm{tan}}^{2}\left(x+y\right)\right)}^{1/2}}\\ x-y=\mathrm{ln}\frac{{C}_{2}\left(1+\mathrm{tan}\left(x+y\right)\right)}{{\left(1+{\mathrm{tan}}^{2}\left(x+y\right)\right)}^{1/2}}\\ {e}^{x-y}=\frac{{C}_{2}\left(1+\frac{\mathrm{sin}\left(x+y\right)}{\mathrm{cos}\left(x+y\right)}\right)}{{\left(1+\frac{{\mathrm{sin}}^{2}\left(x+y\right)}{{\mathrm{cos}}^{2}\left(x+y\right)}\right)}^{1/2}}\end{array}$

$\begin{array}{c}{e}^{x-y}=\frac{{C}_{2}\left(\mathrm{cos}\left(x+y\right)+\mathrm{sin}\left(x+y\right)\right)}{{\left({\mathrm{cos}}^{2}\left(x+y\right)+{\mathrm{sin}}^{2}\left(x+y\right)\right)}^{1/2}}\\ {e}^{x-y}={C}_{2}\left(\mathrm{cos}\left(x+y\right)+\mathrm{sin}\left(x+y\right)\right)\\ \frac{{e}^{x-y}}{{C}_{2}}=\left(\mathrm{cos}\left(x+y\right)+\mathrm{sin}\left(x+y\right)\right)\\ \mathrm{cos}\left(x+y\right)+\mathrm{sin}\left(x+y\right)=C{e}^{x-y}\end{array}$

Where, $C=\frac{1}{{C}_{2}}$ , is an arbitrary constant.

Hence, the solution of the given equation ${\mathbit{c}}{\mathbit{o}}{\mathbit{s}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}{\mathbit{d}}{\mathbit{y}}{\mathbf{=}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}{\mathbit{d}}{\mathbit{x}}$ is ${\mathbit{c}}{\mathbit{o}}{\mathbit{s}}\left(x+y\right){\mathbf{+}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\left(x+y\right){\mathbf{=}}{\mathbit{C}}{{\mathbit{e}}}^{\mathbf{x}\mathbf{-}\mathbf{y}}$.

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