Question

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x \u003e -1, y(0) = 5

Accepted Answer

Rewrite the given differential equation in the standard linear form $$\displaystyle \frac{dy}{dx} + P(x)y = Q(x). $$Start by dividing both sides of the equation by $$(x+1) to $$isolate $$\frac{dy}{dx}$$: $$\displaystyle \frac{dy}{dx} - \frac{2(x^2 + x)}{x+1} y = \frac{e^{x^2}}{(x+1)^2}.$$ Simplify the coefficient of $$y by $$factoring and reducing the expression $$\frac{2(x^2$ + x)}{x+1}. $$Notice that $$x^2 + x = x(x+1$$)$$, so this becomes $$\frac{2x(x+1)}{x+1} = 2x$$ for x$$ \u003e -1$$. Now the equation is $$\displaystyle \frac{dy}{dx} - 2x y = \frac{$$e^{x^2}$$}{$$(x+1)^2$$}$$. Identify P(x) = -2x$ and Q(x$$) = \frac{$$e^{x^2}$$}{$$(x+1)^2$$}$$. Find the integrating factor $$\mu(x)$$ using the formula $$\displaystyle \mu(x) = $$e^{\int P(x) dx}$$ = $$e^{\int -2x dx}$$ = $$e^{-x^2}. $$Multiply both sides of the differential equation by $$\mu(x) to $$write the left side as the derivative of a product: $$\displaystyle e^{-x^2} \frac{dy}{dx} - 2x e^{-x^2} y = \frac{e^{x^2}}{(x+1)^2} e^{-x^2} \implies \frac{d}{dx} \left( y e^{-x^2} \right) = \frac{1}{(x+1)^2}.$$

Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.

(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5

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Key Concepts

Initial Value Problems (IVPs)

First-Order Linear Differential Equations

Integrating Factor Method

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5