Solve the differential and initial value problem:
x(dy/dx) + 1 = y^2
y(1)=0
I tried using bernoulli and it didn't quite work.
xy' + 1 = y^2
xy' = y^2-1
y' = (y^2-1)/x
dy/(y^2-1) = dx/x
arctanh(y) = log(x)
or,
log (1-y)/(1+y) = 2log(x)+c
and you can massage that into exponentials and wind up with
y = 1-e^(2cx^2) / 1+e^(2cx^2)
To solve the given differential equation and initial value problem, we can use an integrating factor. Let's go step by step.
Given differential equation: x(dy/dx) + 1 = y^2
Step 1: Rewrite the equation in standard form
Start by rearranging the equation to separate the terms with y and the terms with x:
x(dy/dx) = y^2 - 1
Step 2: Bring everything involving y to one side
Divide both sides of the equation by y^2 - 1:
x(dy/dx) / (y^2 - 1) = 1
Step 3: Identify a suitable integrating factor
From here, we need to find an integrating factor that makes the left-hand side of the equation an exact derivative of a single function with respect to x. In this case, let's try to use the integrating factor μ(x):
μ(x) = exp(x)
Step 4: Multiply the integrating factor with the equation
Multiply both sides of the equation by μ(x):
exp(x) * x(dy/dx) / (y^2 - 1) = exp(x)
After multiplying, you will get:
(exp(x) * x(dy/dx)) / (y^2 - 1) = exp(x)
Step 5: Apply the product rule on the left-hand side
Apply the product rule of differentiation on the left-hand side:
d/dx[exp(x) * x(dy/dx)] / (y^2 - 1) = exp(x)
Differentiate the product using the product rule:
exp(x) * dy/dx + x * (d/dx(dy/dx)) = exp(x)
Step 6: Simplify and rearrange the equation
Rearrange the equation to solve for d/dx(dy/dx):
dy/dx = (exp(x) - x) / (x * (y^2 - 1))
Step 7: Separate the variables
Rewrite the equation to separate the variables:
(y^2 - 1) / (exp(x) - x) dy = dx / x
Step 8: Integrate both sides
Integrate both sides of the equation:
∫ (y^2 - 1) / (exp(x) - x) dy = ∫ dx / x
Integrating the left-hand side requires some manipulations, so let's break it down further.
Considering the numerator:
∫ (y^2 - 1) dy = ∫ y^2 dy - ∫ dy
= (1/3)y^3 - y + C1
Considering the denominator:
∫ (exp(x) - x) dx
= ∫ exp(x) dx - ∫ x dx
= exp(x) - (1/2)x^2 + C2
Combining both results, we have:
(1/3)y^3 - y + C1 = exp(x) - (1/2)x^2 + C2
Step 9: Re-arrange the equation and simplify
Rearrange the equation to solve for y:
(1/3)y^3 - y = exp(x) - (1/2)x^2 + C
Where C = C2 - C1 is the constant of integration.
Step 10: Apply the initial condition
Now we can apply the initial condition y(1) = 0. Substituting x = 1 and y = 0 into the equation, we get:
(1/3)*0^3 - 0 = exp(1) - (1/2)*1^2 + C
0 = exp(1) - 1/2 + C
C = 1/2 - exp(1)
Therefore, the solution to the given differential equation and initial value problem is:
(1/3)y^3 - y = exp(x) - (1/2)x^2 + (1/2 - exp(1))