Key Concepts and Formulas

• Separable Differential Equations: A differential equation of the form $f(y)dy = g(x)dx$ can be solved by integrating both sides.
• Integration of Rational Functions: Partial fraction decomposition is used to integrate rational functions.
• Standard Integrals: $\int \frac{1}{1+x^2} dx = \arctan(x) + C$

Step-by-Step Solution

Step 1: Separate the variables

The given differential equation is $(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$ We rearrange it to separate the variables $x$ and $y$: $(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y = (2 x^2+2 x+3) \mathrm{d} x$ $\mathrm{d} y = \frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} \mathrm{d} x$

Step 2: Factor the denominator

We try to factor the denominator $x^4+2 x^3+3 x^2+2 x+2$. Notice that $x^4+2x^3+x^2 + 2x^2+2x+2 = x^2(x^2+2x+1) + 2(x^2+x+1)$. Observe that $x^4+2 x^3+3 x^2+2 x+2 = (x^2+1)^2 + 2x(x^2+1) + x^2 = (x^2+x+1)^2 + 1 - x^2 + 2x = (x^2+x+1)^2 + 1 +2x -x^2 = (x^2+x+1)^2 + (1-x^2) + 2x$ However, we can write $x^4+2 x^3+3 x^2+2 x+2 = (x^2+ax+b)(x^2+cx+d) = x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd$ Comparing coefficients, we have $a+c = 2$, $ac+b+d=3$, $ad+bc=2$, $bd=2$. If $b=d=\sqrt{2}$, then $a+c=2$, $ac+2\sqrt{2}=3$, $\sqrt{2}(a+c)=2$, so $a+c = \sqrt{2}$. Contradiction.

Instead, let's try completing the square with $x^2+1$. $x^4+2 x^3+3 x^2+2 x+2 = (x^2+1)^2 + 2x^3 + x^2 + 2x + 1 = (x^2+1)^2 + 2x(x^2+1)+x^2+1 = (x^2+x+1)^2+1-x^2+2x$ Let's check $(x^2+x+1)^2 + (x-1)^2 = x^4+x^2+1 + 2x^3 + 2x^2 + 2x + x^2 -2x + 1= x^4+2x^3+4x^2+2$.

Notice that $x^4+2x^3+3x^2+2x+2 = (x^2+x)^2 + 2(x^2+x) + x^2+2 = (x^2+x+1)^2 + 1 - x^2+2x+1 -1+2 = (x^2+x+1)^2 -(x-1)^2+3$ Let's try factoring as $(x^2+1)(x^2+2x+2) + x^2+2x+2 + 2x^3+x^2+2x = (x^2+1)(x^2+2x+2) = x^4+2x^3+2x^2+x^2+2x+2 = x^4+2x^3+3x^2+2x+2$. Thus, we have $x^4+2x^3+3x^2+2x+2 = (x^2+1)(x^2+2x+2) = (x^2+1)((x+1)^2+1)$

Step 3: Rewrite the integral

Thus, $\mathrm{d} y = \frac{2 x^2+2 x+3}{(x^2+1)(x^2+2 x+2)} \mathrm{d} x = \frac{2 x^2+2 x+3}{(x^2+1)((x+1)^2+1)} \mathrm{d} x$

Step 4: Partial Fraction Decomposition

We decompose the rational function into partial fractions: $\frac{2 x^2+2 x+3}{(x^2+1)(x^2+2 x+2)} = \frac{A x+B}{x^2+1} + \frac{C x+D}{x^2+2 x+2}$ $2 x^2+2 x+3 = (A x+B)(x^2+2 x+2) + (C x+D)(x^2+1)$ $2 x^2+2 x+3 = A x^3+2 A x^2+2 A x+B x^2+2 B x+2 B+C x^3+C x+D x^2+D$ $2 x^2+2 x+3 = (A+C) x^3+(2 A+B+D) x^2+(2 A+2 B+C) x+(2 B+D)$ Comparing coefficients: \begin{align*} A+C &= 0 \ 2 A+B+D &= 2 \ 2 A+2 B+C &= 2 \ 2 B+D &= 3 \end{align*} From the first equation, $C=-A$. From the third equation, $2A+2B-A = 2$, so $A+2B=2$. From the fourth equation, $D=3-2B$. Substituting into the second equation, $2A+B+3-2B = 2$, so $2A-B = -1$. Now we have $A+2B = 2$ and $2A-B = -1$. Multiply the second equation by 2: $4A-2B=-2$. Adding to the first equation, $5A=0$, so $A=0$. Then $C=0$. Then $2B=2$, so $B=1$. Then $D=3-2(1)=1$. Thus, $\frac{2 x^2+2 x+3}{(x^2+1)(x^2+2 x+2)} = \frac{1}{x^2+1} + \frac{1}{x^2+2 x+2} = \frac{1}{x^2+1} + \frac{1}{(x+1)^2+1}$

Step 5: Integrate both sides

$\int \mathrm{d} y = \int \left(\frac{1}{x^2+1} + \frac{1}{(x+1)^2+1}\right) \mathrm{d} x$ $y = \arctan(x) + \arctan(x+1) + K$

Step 6: Apply the initial condition

Given that $y(-1)=-\frac{\pi}{4}$, we have $-\frac{\pi}{4} = \arctan(-1) + \arctan(0) + K$ $-\frac{\pi}{4} = -\frac{\pi}{4} + 0 + K$ $K = 0$ Thus, $y(x) = \arctan(x) + \arctan(x+1)$.

Step 7: Find y(0)

We want to find $y(0)$: $y(0) = \arctan(0) + \arctan(1) = 0 + \frac{\pi}{4} = \frac{\pi}{4}$

Step 8: Re-examine the factoring and the solution

Let's re-examine the original equation and the factorization: $\mathrm{d} y = \frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} \mathrm{d} x$ $\mathrm{d} y = \frac{2 x^2+2 x+3}{(x^2+1)(x^2+2 x+2)} \mathrm{d} x$ $\frac{2 x^2+2 x+3}{(x^2+1)(x^2+2 x+2)} = \frac{A x+B}{x^2+1} + \frac{C x+D}{x^2+2 x+2}$ The correct partial fraction decomposition is: $\frac{1}{x^2+1} + \frac{1}{x^2+2x+2} = \frac{1}{x^2+1} + \frac{1}{(x+1)^2+1}$ $y(x) = \arctan(x) + \arctan(x+1) + K$ Using $y(-1) = -\frac{\pi}{4}$: $-\frac{\pi}{4} = \arctan(-1) + \arctan(0) + K$ $-\frac{\pi}{4} = -\frac{\pi}{4} + 0 + K$ $K=0$ $y(x) = \arctan(x) + \arctan(x+1)$ $y(0) = \arctan(0) + \arctan(1) = 0 + \frac{\pi}{4} = \frac{\pi}{4}$

There must be an error in the problem statement or the provided answer. Let's rework the problem, aiming for an answer of $-\frac{\pi}{12}$.

Suppose the problem meant $y(1) = -\frac{\pi}{4}$. Then $-\frac{\pi}{4} = \arctan(1) + \arctan(2) + K$ $-\frac{\pi}{4} = \frac{\pi}{4} + \arctan(2) + K$ $K = -\frac{\pi}{2} - \arctan(2)$ $y(x) = \arctan(x) + \arctan(x+1) - \frac{\pi}{2} - \arctan(2)$ $y(0) = \arctan(0) + \arctan(1) - \frac{\pi}{2} - \arctan(2) = 0 + \frac{\pi}{4} - \frac{\pi}{2} - \arctan(2) = -\frac{\pi}{4} - \arctan(2)$

Let us assume the initial condition is $y(1) = C$ and we want $y(0) = -\frac{\pi}{12}$. Then $y(0) = \arctan(0) + \arctan(1) + K = \frac{\pi}{4} + K = -\frac{\pi}{12}$, so $K = -\frac{\pi}{12} - \frac{\pi}{4} = -\frac{\pi}{12} - \frac{3\pi}{12} = -\frac{4\pi}{12} = -\frac{\pi}{3}$. Thus $y(x) = \arctan(x) + \arctan(x+1) - \frac{\pi}{3}$. $y(1) = \arctan(1) + \arctan(2) - \frac{\pi}{3} = \frac{\pi}{4} + \arctan(2) - \frac{\pi}{3} = \arctan(2) - \frac{\pi}{12}$.

However, with the given initial condition $y(-1) = -\frac{\pi}{4}$, we found $y(0) = \frac{\pi}{4}$. Let's look for a different error.

The integral is $y = \arctan(x) + \arctan(x+1) + C$. If $y(-1) = -\frac{\pi}{4}$, then $-\frac{\pi}{4} = \arctan(-1) + \arctan(0) + C = -\frac{\pi}{4} + 0 + C$, so $C=0$. Then $y(x) = \arctan(x) + \arctan(x+1)$, so $y(0) = \arctan(0) + \arctan(1) = 0 + \frac{\pi}{4} = \frac{\pi}{4}$.

Let's assume the initial condition is $y(-2) = -\frac{\pi}{4}$. Then $y(-2) = \arctan(-2) + \arctan(-1) + C = -\frac{\pi}{4}$, so $\arctan(-2) - \frac{\pi}{4} + C = -\frac{\pi}{4}$, so $C = -\arctan(-2) = \arctan(2)$. Then $y(x) = \arctan(x) + \arctan(x+1) + \arctan(2)$. $y(0) = \arctan(0) + \arctan(1) + \arctan(2) = \frac{\pi}{4} + \arctan(2)$.

Let's assume $y(0) = -\frac{\pi}{12}$. Then $\arctan(0) + \arctan(1) + C = -\frac{\pi}{12}$, so $0 + \frac{\pi}{4} + C = -\frac{\pi}{12}$, so $C = -\frac{\pi}{12} - \frac{\pi}{4} = -\frac{\pi}{12} - \frac{3\pi}{12} = -\frac{4\pi}{12} = -\frac{\pi}{3}$. Then $y(x) = \arctan(x) + \arctan(x+1) - \frac{\pi}{3}$. If $y(-1) = -\frac{\pi}{4}$, then $\arctan(-1) + \arctan(0) - \frac{\pi}{3} = -\frac{\pi}{4}$, so $-\frac{\pi}{4} + 0 - \frac{\pi}{3} = -\frac{\pi}{4} - \frac{\pi}{3} \neq -\frac{\pi}{4}$.

It appears there is an error in the question itself. Given the differential equation and the initial condition $y(-1) = -\frac{\pi}{4}$, the correct value for $y(0)$ is $\frac{\pi}{4}$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs during partial fraction decomposition and integration.
• Checking Factorization: Always verify your factorization to avoid errors.
• arctan range: The range of arctan is $(-\pi/2, \pi/2)$.

Summary

We separated the variables in the given differential equation, factored the denominator, performed partial fraction decomposition, integrated both sides, and applied the initial condition to find the constant of integration. Based on the given differential equation and initial condition, $y(0) = \frac{\pi}{4}$. However, the provided correct answer is $-\frac{\pi}{12}$, suggesting there is an error in the problem statement or the given answer. Assuming the problem statement is correct, the correct value for $y(0)$ is $\frac{\pi}{4}$.

Final Answer

The final answer is \boxed{\frac{\pi}{4}}, which corresponds to option (D).