solve differential equation dy/dx=7y^2/sqrtx, y(1)=1/36
y(x)=
AAAaannndd the bot gets it wrong yet again!
dy/dx =7y^2/√x
dy/y^2 = 7/√x dx
-1/y = 14√x + C
since y(1) = 1/36
-36 = 14+C
C = -50
so that gives you
y = 1/(50-14√x)
To solve the differential equation `dy/dx = 7y^2/sqrt(x)`, we need to separate the variables `y` and `x`. We can do this by multiplying both sides by `sqrt(x)/y^2`:
```
y'(x) = 7y^2 / sqrt(x)
sqrt(x) / y^2 dy/dx = 7
```
Now we integrate both sides with respect to `x` and `y`:
```
int sqrt(x) / y^2 dy = 7 int dx
-1/y = 14/3 x^(3/2) + C
```
where `C` is the constant of integration. To find `C`, we use the initial condition `y(1) = 1/36`:
```
-1/(1/36) = 14/3 (1)^(3/2) + C
C = -37/108
```
Therefore, the solution to the differential equation is:
```
-1/y = 14/3 x^(3/2) - 37/108
y(x) = -1 / (14/3 x^(3/2) - 37/108)
y(x) = 108 / (42 x^(3/2) - 37)
```