Question

1. x^2y'-2xy+y'=02. (x-y) dy=ydx​

Answer 1

|x+y-2|+x^2-2xy+y^2=0

|x+y-2|+(x-y)^2 = 0

{x+y-2 = 0

{x-y = 0

x=y

y+y-2 = 0

2y =2

y=1

Answer 2

To solve the given differential equations, let's take them one by one:

1. x^2y' - 2xy + y' = 0

To solve this first-order linear ordinary differential equation, we'll group the y' terms together and factor out y':

x^2y' + y' - 2xy = 0

Now, factor out y' from the left-hand side:

y' (x^2 + 1) - 2xy = 0

Now, we can isolate y' by dividing both sides by (x^2 + 1):

y' = 2xy / (x^2 + 1)

Next, we can separate variables by moving y to one side and dx to the other side:

dy/dx = 2xy / (x^2 + 1)

Now, we can integrate both sides with respect to x:

∫(1/y) dy = ∫(2x / (x^2 + 1)) dx

The integral on the left side is simply ln|y|, and the integral on the right side requires a substitution (u-substitution):

Let u = x^2 + 1, then du = 2x dx

The integral becomes:

∫(2x / (x^2 + 1)) dx = ∫(1/u) du = ln|u| + C = ln|x^2 + 1| + C

So, the solution to the differential equation is:

ln|y| = ln|x^2 + 1| + C

Where C is the constant of integration. We can combine the logarithms and raise them as a power of the base:

|y| = |x^2 + 1| * e^C

The constant of integration, C, can be absorbed into another constant, so we simplify as:

|y| = A * |x^2 + 1|

Where A is the combined constant. Since y can be positive or negative, we can remove the absolute value signs:

y = ±A * (x^2 + 1)

This is the general solution to the differential equation.

2. (x - y) dy = y dx

To solve this separable differential equation, we'll move all y terms to one side and all x terms to the other side:

(x - y) dy - y dx = 0

Now, we can divide both sides by (x - y):

dy/dx = y / (x - y)

Next, we can separate variables by moving y to one side and dx to the other side:

dy / y = dx / (x - y)

Now, we can integrate both sides:

∫(1/y) dy = ∫(1/(x - y)) dx

The integral on the left side is ln|y|, and the integral on the right side requires a substitution:

Let u = x - y, then du = -dy

The integral becomes:

∫(1/(x - y)) dx = -∫(1/u) du = -ln|u| + C = -ln|x - y| + C

So, the solution to the differential equation is:

ln|y| = -ln|x - y| + C

We can combine the logarithms and raise them as a power of the base:

|y| = e^(C) / |x - y|

The constant of integration, C, can be absorbed into another constant, so we simplify as:

|y| = A / |x - y|

Where A is the combined constant. Since y can be positive or negative, we can remove the absolute value signs:

y = ±A / |x - y|

This is the general solution to the differential equation.