Math Problem Statement

The variables x and y satisfy the differential equation dy/dx = x^2 * e^(y + 2x) and it is given that y = 0 when x = 0. Solve the differential equation and obtain an expression for y in terms of x.

Solution

The differential equation given is:

dydx=x2ey+2x\frac{dy}{dx} = x^2 e^{y + 2x}

with the initial condition y=0y = 0 when x=0x = 0. The goal is to solve this differential equation and express yy in terms of xx.

Step 1: Rearranging the Equation

The equation can be rearranged to isolate yy and xx terms:

dyey=x2e2xdx\frac{dy}{e^y} = x^2 e^{2x} \, dx

Step 2: Integrate Both Sides

To solve this, integrate both sides separately:

1eydy=x2e2xdx\int \frac{1}{e^y} \, dy = \int x^2 e^{2x} \, dx

For the left-hand side:

1eydy=eydy=ey\int \frac{1}{e^y} \, dy = \int e^{-y} \, dy = -e^{-y}

For the right-hand side, integration by parts or using substitution methods might be required. Let's proceed with that step.

Right-hand Side Integration:

To solve x2e2xdx\int x^2 e^{2x} \, dx, we can use the method of integration by parts. Let:

  • u=x2u = x^2 and dv=e2xdxdv = e^{2x} \, dx
  • Thus, du=2xdxdu = 2x \, dx and v=12e2xv = \frac{1}{2} e^{2x}

Now applying integration by parts:

x2e2xdx=x212e2x2x12e2xdx\int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int 2x \cdot \frac{1}{2} e^{2x} \, dx

=x22e2xxe2xdx= \frac{x^2}{2} e^{2x} - \int x e^{2x} \, dx

We again apply integration by parts on xe2xdx\int x e^{2x} \, dx:

  • u=xu = x and dv=e2xdxdv = e^{2x} \, dx
  • Thus, du=dxdu = dx and v=12e2xv = \frac{1}{2} e^{2x}

xe2xdx=x12e2x12e2xdx=x2e2x14e2x\int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x}

Thus:

x2e2xdx=x22e2x(x2e2x14e2x)=(x22x2+14)e2x\int x^2 e^{2x} \, dx = \frac{x^2}{2} e^{2x} - \left(\frac{x}{2} e^{2x} - \frac{1}{4} e^{2x}\right) = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x}

Step 3: Combining and Applying the Initial Condition

Now we have:

ey=(x22x2+14)e2x+C-e^{-y} = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} + C

Applying the initial condition x=0x = 0 and y=0y = 0:

e0=(02202+14)e0+C    1=14+C    C=54-e^{0} = \left(\frac{0^2}{2} - \frac{0}{2} + \frac{1}{4}\right) e^{0} + C \implies -1 = \frac{1}{4} + C \implies C = -\frac{5}{4}

Thus, the solution is:

ey=(x22x2+14)e2x54-e^{-y} = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} - \frac{5}{4}

Step 4: Solving for yy

To find yy, isolate eye^{-y}:

ey=((x22x2+14)e2x54)e^{-y} = -\left(\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} - \frac{5}{4}\right)

Taking the natural logarithm on both sides to find yy:

y=ln((x22x2+14)e2x54)y = -\ln\left(\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} - \frac{5}{4}\right)

This is the final expression for yy in terms of xx.

Would you like more details on the steps, or do you have any questions?


Here are five related questions:

  1. How does the integration by parts method work, and when should it be used?
  2. Why is an integrating factor not applicable in solving this differential equation?
  3. How do we interpret the natural logarithm of an expression in the context of differential equations?
  4. What would happen if a different initial condition were given instead of y=0y = 0 at x=0x = 0?
  5. Could we solve this differential equation numerically if we couldn't find a closed form?

Tip: In solving differential equations, always check if you can separate variables before resorting to more complex techniques like integration by parts or substitution.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Separable Equations
Integration by Parts
Initial Value Problems

Formulas

dy/dx = x^2 * e^(y + 2x)
Integration by parts: ∫u dv = u v - ∫v du
General solution of first-order differential equations

Theorems

Fundamental Theorem of Calculus
Integration by Parts Technique

Suitable Grade Level

Undergraduate - Calculus II