Question

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

Short answer

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

Step-by-step solution

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.

Analyzing the given statement

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

$\cos (x+y)dy=\sin (x+y)dx\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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

Rewriting the equation (1) as,

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

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

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,

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

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

Find the solution of equation  dzdx=1+tanz

Separate the variables in equation (4),

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

Integrating both sides,

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

Put tan z = t

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

Thus, equation (5) becomes,

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

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 $\left(1+t\right)\left(1+t^2\right)$

$1=A\left(1+t^2\right)+Bt\left(1+t\right)+C_1\left(1+t\right)\cdot \cdot \cdot \cdot \cdot \cdot \left(7\right)$

Put t= -1

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

Equating coefficients in equation (7) of:

t²)$A+B=0\cdot \cdot \cdot \cdot \cdot \cdot \left(8\right)$

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,

$\begin{array}{c}\int dx=\frac{1}{2}\int \frac{dt}{\left(1+t\right)}-\frac{1}{2}\int \frac{t}{1+t^2}dt+\frac{1}{2}\int \frac{dt}{1+t^2}\\ \int dx=\frac{1}{2}\int \frac{dt}{\left(1+t\right)}-\frac{1}{4}\int \frac{2t}{1+t^2}dt+\frac{1}{2}\int \frac{dt}{1+t^2}\\ x=\frac{1}{2}\ln \left|1+t\right|-\frac{1}{4}\ln \left|1+t^2\right|+\frac{1}{2}{\tan }^{-1}t+\ln C_2\end{array}$

Where, C₂ is the constant of integration.

Put t = tan z

$\begin{array}{c}x=\frac{1}{2}\ln \left|1+\tan z\right|-\frac{1}{4}\ln \left|1+{\tan }^{2}z\right|+\frac{1}{2}{\tan }^{-1}\left(\tan z\right)+\ln C_2\\ x=\frac{1}{2}\ln \left|1+\tan z\right|-\frac{1}{4}\ln \left|1+{\tan }^{2}z\right|+\frac{1}{2}z+\ln C_2\\ 2x=\ln \left|1+\tan z\right|-\frac{1}{2}\ln \left|1+{\tan }^{2}z\right|+z+\ln C_2\\ 2x-z=\ln \left|1+\tan z\right|-\ln {\left(1+{\tan }^{2}z\right)}^{1/2}+\ln C_2\end{array}$

$\begin{array}{c}2x-z=\ln \frac{\left(1+\tan z\right)}{{\left(1+{\tan }^{2}z\right)}^{1/2}}+\ln C_2\\ 2x-z=\ln \frac{C_2\left(1+\tan z\right)}{{\left(1+{\tan }^{2}z\right)}^{1/2}}\end{array}$

Put z = x + y

$\begin{array}{c}2x-\left(x+y\right)=\ln \frac{C_2\left(1+\tan \left(x+y\right)\right)}{{\left(1+{\tan }^2\left(x+y\right)\right)}^{1/2}}\\ 2x-x-y=\ln \frac{C_2\left(1+\tan \left(x+y\right)\right)}{{\left(1+{\tan }^2\left(x+y\right)\right)}^{1/2}}\\ x-y=\ln \frac{C_2\left(1+\tan \left(x+y\right)\right)}{{\left(1+{\tan }^2\left(x+y\right)\right)}^{1/2}}\\ e^{x-y}=\frac{C_2\left(1+\frac{\sin \left(x+y\right)}{\cos \left(x+y\right)}\right)}{{\left(1+\frac{{\sin }^2\left(x+y\right)}{{\cos }^2\left(x+y\right)}\right)}^{1/2}}\end{array}$

$\begin{array}{c}{e}^{x-y}=\frac{C_2\left(\cos \left(x+y\right)+\sin \left(x+y\right)\right)}{{\left({\cos }^2\left(x+y\right)+{\sin }^2\left(x+y\right)\right)}^{1/2}}\\ e^{x-y}=C_2\left(\cos \left(x+y\right)+\sin \left(x+y\right)\right)\\ \frac{e^{x-y}}{C_2}=\left(\cos \left(x+y\right)+\sin \left(x+y\right)\right)\\ \cos \left(x+y\right)+\sin \left(x+y\right)=C{e}^{x-y}\end{array}$

Where, $C=\frac{1}{C_2}$, is an arbitrary constant.

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