Math Problem Statement

y´=1+x+y+xy

Solution

The given differential equation is:

y=1+x+y+xyy' = 1 + x + y + xy

This can be rewritten as:

dydx=1+x+y+xy\frac{dy}{dx} = 1 + x + y + xy

This equation is not separable or exact in its current form, but it can be simplified using a substitution technique. One common substitution is to use:

v=y+1v = y + 1

Differentiating both sides with respect to xx, we get:

dvdx=dydx\frac{dv}{dx} = \frac{dy}{dx}

Now substitute y=v1y = v - 1 into the original equation:

dvdx=1+x+(v1)+x(v1)\frac{dv}{dx} = 1 + x + (v - 1) + x(v - 1)

Simplifying:

dvdx=1+x+v1+xvx\frac{dv}{dx} = 1 + x + v - 1 + xv - x

dvdx=v+xv\frac{dv}{dx} = v + xv

Factor out vv:

dvdx=v(1+x)\frac{dv}{dx} = v(1 + x)

This is a separable equation now. We can rewrite it as:

dvv=(1+x)dx\frac{dv}{v} = (1 + x) dx

Integrating both sides:

lnv=x22+x+C\ln|v| = \frac{x^2}{2} + x + C

Exponentiating both sides:

v=ex22+x+C=eCex22+xv = e^{\frac{x^2}{2} + x + C} = e^C e^{\frac{x^2}{2} + x}

Let eC=C1e^C = C_1, then:

v=C1ex22+xv = C_1 e^{\frac{x^2}{2} + x}

Since v=y+1v = y + 1, we have:

y+1=C1ex22+xy + 1 = C_1 e^{\frac{x^2}{2} + x}

Finally, solving for yy:

y=C1ex22+x1y = C_1 e^{\frac{x^2}{2} + x} - 1

Thus, the general solution is:

y=C1ex22+x1y = C_1 e^{\frac{x^2}{2} + x} - 1

Would you like more details on any step or clarification?

Related Questions:

  1. How do you solve separable differential equations in general?
  2. What are the conditions for a differential equation to be exact?
  3. Can other substitution methods simplify this differential equation further?
  4. How do boundary or initial conditions affect the constant C1C_1?
  5. What are common techniques to solve first-order differential equations?

Tip:

Always check if a differential equation is separable, exact, or can be transformed via substitution for easier solving methods.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Substitution Method
Separable Equations

Formulas

y' = 1 + x + y + xy
v = y + 1
\frac{dv}{dx} = v(1 + x)
\ln|v| = \frac{x^2}{2} + x + C
v = C_1 e^{\frac{x^2}{2} + x}
y = C_1 e^{\frac{x^2}{2} + x} - 1

Theorems

Separable Differential Equations
Substitution in Differential Equations

Suitable Grade Level

College Level - Calculus II or Differential Equations