We start with the differential equation:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y}{x - y}

To solve this, we'll use the method of substitution.

This substitution implies that y = v x . By expressing y in terms of v and x , we can transform the differential equation into a form that is easier to solve.

Using the product rule, we have:

\frac{\mathrm{d}y}{\mathrm{d}x}=v+x\frac{\mathrm{d}v}{\mathrm{d}x}

Substitute y = v x and \frac{\mathrm{d}y}{\mathrm{d}x} = v + x \frac{\mathrm{d}v}{\mathrm{d}x} into the differential equation :

v + x \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{x + v x}{x - v x} = \frac{1 + v}{1 - v}

Rearrange the equation to isolate \frac{\mathrm{d}v}{\mathrm{d}x}:

x \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v (1 - v)}{1 - v} = \frac{1 + v^2}{1 - v}

Now, separate the variables v and x :

\frac{1 - v}{1 + v^2} \, \mathrm{d}v = \frac{\mathrm{d}x}{x}

Integrate both sides of the equation:

\int \frac{1 - v}{1 + v^2} \, \mathrm{d}v = \int \frac{\mathrm{d}x}{x}

The left side can be split into two separate integrals:

\int \frac{1}{1 + v^2} \, \mathrm{d}v - \int \frac{v}{1 + v^2} \, \mathrm{d}v

The first integral, \int \frac{1}{1 + v^2} , \mathrm{d}v, is the standard arctangent integral:

\int \frac{1}{1 + v^2} \, \mathrm{d}v = \arctan(v)

The second integral, \int \frac{v}{1 + v^2} , \mathrm{d}v, can be solved using the substitution u = 1 + v^2, which gives \mathrm{d}u = 2v , \mathrm{d}v:

\int \frac{v}{1 + v^2} \, \mathrm{d}v = \frac{1}{2} \ln(1 + v^2)

Thus, the left side becomes:

\arctan(v) - \frac{1}{2} \ln(1 + v^2)

The right side is a simple logarithmic integral:

\int \frac{\mathrm{d}x}{x} = \ln|x|

Putting it all together, we have:

\arctan(v) - \frac{1}{2} \ln(1 + v^2) = \ln|x| + C

Replace v with \frac{y}{x} :

\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C

Simplify the expression:

\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2 + y^2) = \ln|x| + C

This equation represents the general solution to the differential equation in an implicit form.

Substitute the initial condition into the equation to find C :

\arctan\left(\frac{0}{1}\right) - \frac{1}{2} \ln(1^2 + 0^2) = \ln|1| + C

0 - \frac{1}{2} \ln(1) = 0 + C \implies C = 0

The solution to the differential equation, incorporating the initial condition, is:

\boxed{\arctan\left(\frac{y}{x}\right)-\frac12\ln(x^2+y^2)=\ln|x|}

Problem Bank - Differential Equations

Problem Bank - Differential Equations