What is a solution to the differential equation #dy/dx=y^2+1# with y(1)=0?

Question

Paisley Johnson · Accepted Answer

#y = tan(x-1)#

#dy/dx=y^2+1#

grouping variables

#dy/(y^2+1) = dx#

Integrating both sides

#arctan(y) = x + C#

then

#y = tan(x+C)#

Now, observing the initial condition

#y(1)=tan(1+C)=0# so

#1+C=0+kpi, k=pm1,pm2,cdots#

and concluding

#y = tan(x-1)#

Emily Ammerman · Answer

undefined The solution to the differential equation $ \frac{dy}{dx} = y^2 + 1 $ with the initial condition $ y(1) = 0 $ is $ y(x) = \tan(x) $.

Aurora Alvarado · Answer

undefined To solve the given initial value problem, we can use the method of separation of variables.

Given the differential equation $ \frac{dy}{dx} = y^2 + 1 $, we separate variables:

$$ \frac{dy}{y^2 + 1} = dx $$

Now, integrate both sides:

$$ \int \frac{dy}{y^2 + 1} = \int dx $$

The integral on the left side can be solved using the arctangent function:

$$ \arctan(y) = x + C $$

Where $ C $ is the constant of integration.

Now, to find the value of the constant $ C $, we use the initial condition $ y(1) = 0 $:

$$ \arctan(0) = 1 + C $$
$$ 0 = C + \frac{\pi}{4} $$
$$ C = -\frac{\pi}{4} $$

Therefore, the particular solution to the given differential equation with the initial condition $ y(1) = 0 $ is:

$$ \arctan(y) = x - \frac{\pi}{4} $$

Or, equivalently:

$$ y = \tan(x - \frac{\pi}{4}) $$