Question

Solve \(y^{\prime} \sqrt{1+x+y}=x+y-1\)

Short answer

Question: Given the differential equation \(y^{\prime} \sqrt{1+x+y}=x+y-1\), find a function y(x) that satisfies the given equation. Answer: We have found an implicit solution for the given differential equation as follows: $$\int \frac{dy}{(y-1+\sqrt{1+x+y})}=\int \frac{dx}{\sqrt{1+x+y}}+C,$$ where C is the integration constant. Unfortunately, it is not possible to find an explicit solution for y(x) due to the complexity of the integrals involved.

Step-by-step solution

Check for separability and find an integrating factor

The given differential equation is \(y^{\prime} \sqrt{1+x+y}=x+y-1\). It is not directly separable, so we will look for an integrating factor to simplify it. Let's rewrite the equation in terms of dy and dx: $$\frac{dy}{dx} \sqrt{1+x+y}=x+y-1.$$

Finding a potential integrating factor

We will try the integrating factor \(μ(x, y) = \frac{1}{\sqrt{1+x+y}}\). Then, let's multiply both sides of the equation by this integrating factor: $$\frac{1}{\sqrt{1+x+y}}\frac{dy}{dx} \sqrt{1+x+y}=\frac{1}{\sqrt{1+x+y}}(x+y-1).$$ The left-hand side simplifies to: $$\frac{dy}{dx}=(x+y-1)\frac{1}{\sqrt{1+x+y}}.$$

Check for separability

Now the differential equation is in the form: $$\frac{dy}{dx}=(x+y-1)\frac{1}{\sqrt{1+x+y}}.$$ We can rewrite this as: $$\frac{dy}{(y-1+\sqrt{1+x+y})}=\frac{dx}{\sqrt{1+x+y}}.$$ Now, the differential equation is separable.

Integrate both sides

We can now integrate both sides of the equation with respect to x and y: $$\int \frac{dy}{(y-1+\sqrt{1+x+y})}=\int \frac{dx}{\sqrt{1+x+y}}.$$

Solve for y(x)

At this point, we don't have a closed-form solution for the integrals. Therefore, we can write the implicit solution for the given differential equation as follows: $$\int \frac{dy}{(y-1+\sqrt{1+x+y})}=\int \frac{dx}{\sqrt{1+x+y}}+C,$$ where C is the integration constant. This is the general solution to the given differential equation. Unfortunately, it is not possible to find an explicit solution for y(x) due to the complexity of the integrals involved.