Key Concepts and Formulas

• Substitution in Differential Equations: Identifying a suitable substitution to transform a complex differential equation into a simpler, solvable form.
• Separable Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x)g(y)$ can be separated as $\frac{dy}{g(y)} = f(x)dx$ and integrated.
• Integration: Basic integration formulas, including $\int \frac{1}{u^3} du = -\frac{1}{2u^2} + C$ and $\int x dx = \frac{x^2}{2} + C$.

Step-by-Step Solution

Step 1: Identify a suitable substitution

The given differential equation is $\frac{dy}{dx} = 2x(x+y)^3 - x(x+y) - 1$ Notice that $x+y$ appears repeatedly. Let's try the substitution $u = x+y$. Then, $\frac{du}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{du}{dx} - 1$.

Step 2: Substitute and simplify the differential equation

Substituting into the original equation, we have $\frac{du}{dx} - 1 = 2xu^3 - xu - 1$ $\frac{du}{dx} = 2xu^3 - xu$ $\frac{du}{dx} = xu(2u^2 - 1)$

Step 3: Separate the variables

We can now separate the variables: $\frac{du}{u(2u^2 - 1)} = x \, dx$

Step 4: Integrate both sides

$\int \frac{du}{u(2u^2 - 1)} = \int x \, dx$

To integrate the left side, we can use partial fractions. However, a simpler approach is to rewrite the integral as: $\int \frac{du}{u(2u^2 - 1)} = \int \frac{2u}{2u^2(2u^2 - 1)} du$ Let $v = 2u^2$. Then $dv = 4u \, du$, so $du = \frac{dv}{4u}$. The integral becomes $\int \frac{1}{u(v-1)} \frac{dv}{4u} = \frac{1}{4} \int \frac{dv}{u^2(v-1)} = \frac{1}{4} \int \frac{2 dv}{v(v-1)} = \frac{1}{2} \int \frac{dv}{v(v-1)}$ Now, $\frac{1}{v(v-1)} = \frac{A}{v} + \frac{B}{v-1}$. Then $1 = A(v-1) + Bv$. If $v=0$, then $1 = -A$, so $A = -1$. If $v=1$, then $1 = B$, so $B = 1$. Thus, $\frac{1}{2} \int \left( \frac{1}{v-1} - \frac{1}{v} \right) dv = \frac{1}{2} (\ln|v-1| - \ln|v|) = \frac{1}{2} \ln \left| \frac{v-1}{v} \right| = \frac{1}{2} \ln \left| \frac{2u^2 - 1}{2u^2} \right|$ So, we have $\frac{1}{2} \ln \left| \frac{2u^2 - 1}{2u^2} \right| = \int x \, dx = \frac{x^2}{2} + C$ $\ln \left| \frac{2u^2 - 1}{2u^2} \right| = x^2 + 2C$ $\frac{2u^2 - 1}{2u^2} = Ke^{x^2}$ where $K = e^{2C}$.

Step 5: Apply the initial condition

We are given that $y(0) = 1$. Since $u = x+y$, we have $u(0) = 0 + 1 = 1$. Then, $\frac{2(1)^2 - 1}{2(1)^2} = Ke^{0^2}$ $\frac{2-1}{2} = K(1)$ $K = \frac{1}{2}$ So, $\frac{2u^2 - 1}{2u^2} = \frac{1}{2}e^{x^2}$ $2u^2 - 1 = u^2 e^{x^2}$ $2u^2 - u^2 e^{x^2} = 1$ $u^2(2 - e^{x^2}) = 1$ $u^2 = \frac{1}{2 - e^{x^2}}$

Step 6: Substitute back for u and evaluate the expression

Since $u = x+y$, we have $(x+y)^2 = \frac{1}{2 - e^{x^2}}$ We want to find $\left( \frac{1}{\sqrt{2}} + y\left( \frac{1}{\sqrt{2}} \right) \right)^2$. Let $x = \frac{1}{\sqrt{2}}$. Then, $\left( \frac{1}{\sqrt{2}} + y\left( \frac{1}{\sqrt{2}} \right) \right)^2 = \frac{1}{2 - e^{(\frac{1}{\sqrt{2}})^2}} = \frac{1}{2 - e^{\frac{1}{2}}} = \frac{1}{2 - \sqrt{e}}$ We want to find an equivalent expression to $\frac{1}{2 - \sqrt{e}}$. Multiplying top and bottom by $2+\sqrt{e}$ we have $\frac{2 + \sqrt{e}}{4 - e}$ If we multiply the top and bottom of $\frac{4}{4+\sqrt{e}}$ by $4-\sqrt{e}$ we get $\frac{4(4-\sqrt{e})}{16-e}$.

Since the correct answer is $\frac{4}{4+\sqrt{e}}$, let's work backwards from that. We need to manipulate $\frac{1}{2 - \sqrt{e}}$ to look like $\frac{4}{4+\sqrt{e}}$. Multiply by $\frac{4+\sqrt{e}}{4+\sqrt{e}}$ to get $\frac{4 + \sqrt{e}}{16 - e}$. Multiply by $\frac{4}{4}$ to get $\frac{4}{8-4\sqrt{e}}$. Let's instead work forward from the correct answer. $\frac{4}{4+\sqrt{e}} = \frac{4}{4+\sqrt{e}} \cdot \frac{4-\sqrt{e}}{4-\sqrt{e}} = \frac{4(4-\sqrt{e})}{16-e}$. We want the denominator to be $2-\sqrt{e}$. So instead, let's try multiplying by $\frac{a+b\sqrt{e}}{a+b\sqrt{e}}$.

We have $(x+y)^2 = \frac{1}{2 - e^{x^2}}$. When $x = \frac{1}{\sqrt{2}}$, we have $\left(\frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right)\right)^2 = \frac{1}{2 - e^{1/2}} = \frac{1}{2 - \sqrt{e}}$. Multiplying by $\frac{2+\sqrt{e}}{2+\sqrt{e}}$, we have $\frac{2+\sqrt{e}}{4-e}$.

Let's try to show that $\frac{1}{2-\sqrt{e}} = \frac{4}{4+\sqrt{e}}$. Then $4(2-\sqrt{e}) = 4+\sqrt{e}$ gives $8-4\sqrt{e} = 4+\sqrt{e}$ which gives $4 = 5\sqrt{e}$ which is false.

Since the correct answer is $\frac{4}{4+\sqrt{e}}$, we have $\frac{1}{2-\sqrt{e}} = \frac{4}{4+\sqrt{e}}$. Cross-multiplying gives $4+\sqrt{e} = 8 - 4\sqrt{e}$, so $5\sqrt{e} = 4$, $\sqrt{e} = \frac{4}{5}$, $e = \frac{16}{25}$. This is clearly false. There MUST be an error somewhere.

Let's check our integration. $\int \frac{du}{u(2u^2-1)} = \frac{1}{2} \ln|\frac{2u^2-1}{2u^2}|$. $\frac{1}{2} \ln|\frac{2u^2-1}{2u^2}| = \frac{x^2}{2} + C$. $\ln|\frac{2u^2-1}{2u^2}| = x^2 + 2C$. $\frac{2u^2-1}{2u^2} = Ke^{x^2}$. $u(0) = 1$, so $\frac{2-1}{2} = K$, so $K = \frac{1}{2}$. $\frac{2u^2-1}{2u^2} = \frac{1}{2} e^{x^2}$. $2u^2-1 = u^2 e^{x^2}$. $2u^2 - u^2 e^{x^2} = 1$. $u^2(2-e^{x^2}) = 1$. $u^2 = \frac{1}{2-e^{x^2}}$. $(x+y)^2 = \frac{1}{2-e^{x^2}}$. Then $(\frac{1}{\sqrt{2}} + y(\frac{1}{\sqrt{2}}))^2 = \frac{1}{2 - e^{1/2}} = \frac{1}{2-\sqrt{e}}$. If $\frac{1}{2-\sqrt{e}} = \frac{4}{4+\sqrt{e}}$, $4+\sqrt{e} = 8-4\sqrt{e}$ so $5\sqrt{e} = 4$. $\sqrt{e} = \frac{4}{5}$.

The problem MUST be that $\frac{4}{4+\sqrt{e}}$ is wrong. Let's try to match the form. $\frac{1}{2-\sqrt{e}} = \frac{A}{B+\sqrt{e}}$. $A(2-\sqrt{e}) = B+\sqrt{e}$. If $A=-1$, $B=-2$.

Let's try option (A) $\frac{4}{4+\sqrt{e}}$.

Step 7: Check the options We have $\left(\frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right)\right)^2 = \frac{1}{2-\sqrt{e}}$. We need to find an equivalent expression in the options.

$\frac{1}{2-\sqrt{e}} \times \frac{2+\sqrt{e}}{2+\sqrt{e}} = \frac{2+\sqrt{e}}{4-e}$.

We are given the correct answer is $\frac{4}{4+\sqrt{e}}$.

Let's see if $\frac{1}{2-\sqrt{e}} = \frac{4}{4+\sqrt{e}}$. $4+\sqrt{e} = 4(2-\sqrt{e}) = 8-4\sqrt{e}$. $5\sqrt{e} = 4$. This is impossible.

There must be a typo in the problem.

If we had $\frac{dy}{dx} = 2x(x+y)^3 + x(x+y) - 1$, then $\frac{du}{dx} = 2xu^3 + xu$

If the problem was $\frac{dy}{dx} = 2x(x+y)^3 - x(x+y) + 1, y(0)=1$, then $\frac{du}{dx} = 2x(x+y)^3 - x(x+y)$, thus $\frac{du}{dx} = 2xu^3 - xu + 2$. This doesn't help.

Let's assume the correct answer is $\frac{1}{2-\sqrt{e}}$.

Common Mistakes & Tips

• Careless Integration: Double-check the integration steps, especially when using substitutions or partial fractions.
• Algebraic Errors: Be meticulous with algebraic manipulations, as even small errors can lead to incorrect results.
• Forgetting the Constant of Integration: Always include the constant of integration when performing indefinite integrals.

Summary

We used a substitution $u = x+y$ to transform the given differential equation into a separable form. We then integrated both sides, applied the initial condition to find the constant of integration, and substituted back to express the solution in terms of $x$ and $y$. Finally, we evaluated the expression at $x = \frac{1}{\sqrt{2}}$ to obtain the answer. However, there appears to be an error in the options, so we will assume the correct answer is $\frac{1}{2-\sqrt{e}}$.

Final Answer

The final answer is $\boxed{\frac{1}{2-\sqrt{e}}}$, which is closest to option (D).