The solution of the differential equation x dy/dx + 2y = x² (x ≠ 0) with y(1) = 1, is (1) y = x³/5 +

Q: Wait, why do we have to divide by $x$ at the start? Can't we just integrate directly?

That is a great question! The standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ is a specific structure that allows us to use the Integrating Factor method. Without that $1$ as the coefficient of $\frac{dy}{dx}$, the product rule trick—which is the heart of this method—simply will not work. We must clean the equation first to reveal its true structure.

Q: I got stuck at $e^{2 \ln x}$. How does that become $x^2$?

Ah, the power of logarithms! Remember the property $a \ln b = \ln(b^a)$. So, $2 \ln x$ becomes $\ln(x^2)$. Since the exponential function $e^x$ and the natural logarithm $\ln x$ are inverse functions, $e^{\ln(x^2)}$ simplifies beautifully to just $x^2$. It is one of the most satisfying moments in calculus!

Q: Do I really need the $+ C$ before finding the specific solution?

Absolutely! The general solution $y = \frac{x^2}{4} + \frac{C}{x^2}$ represents an entire family of curves. Without that $+ C$, you are only looking at one specific curve, not the entire solution space. You need the general form first to apply the initial condition $y(1) = 1$ correctly to 'pin' the curve to the right spot.

Question

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2 (x 
eq 0)$ with $y(1) = 1$, is

Options:

(A) $y = \frac{x^3}{5} + \frac{1}{5x^2}$

(B) $y = \frac{4}{5}x^3 + \frac{1}{5x^2}$

(C) $y = \frac{3}{4}x^2 + \frac{1}{4x^2}$

(D) $y = \frac{x^2}{4} + \frac{3}{4x^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

A differential equation is a set of instructions describing how a function changes at every point. We are tasked with solving the equation:

$$x \frac{dy}{dx} + 2y = x^2$$

subject to the initial condition $y(1) = 1$. This is a classic linear differential equation, and we will solve it using the standard algorithmic approach.

### The Cleaning Phase

When we first look at the equation, the coefficient of $\frac{dy}{dx}$ is $x$. In the world of linear differential equations, we require the coefficient of $\frac{dy}{dx}$ to be exactly $1$.

We perform our first act of mathematical hygiene by dividing the entire equation by $x$:

$$\frac{dy}{dx} + \frac{2}{x}y = x$$

This perfectly matches the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. We have successfully identified our components: $P(x) = \frac{2}{x}$ and $Q(x) = x$.

### The Magic of the Integrating Factor

To unlock this equation, we calculate the Integrating Factor ($IF$). The formula is defined as:

$$IF = e^{\int P(x) dx}$$

Substituting our $P(x)$ into this formula, we get:

$$IF = e^{\int \frac{2}{x} dx} = e^{2 \ln x}$$

Using the logarithm property $a \ln b = \ln(b^a)$, we transform $2 \ln x$ into $\ln(x^2)$. Since $e$ and $\ln$ are inverse functions, they cancel out, leaving us with:

$$IF = x^2$$

### The Execution

We multiply our standard equation by this key ($x^2$):

$$x^2 \left( \frac{dy}{dx} + \frac{2}{x}y \right) = x^2 \cdot x$$

This simplifies to:

$$x^2 \frac{dy}{dx} + 2xy = x^3$$

Notice that the left side is the derivative of the product $y \cdot x^2$. We can rewrite the equation as:

$$\frac{d}{dx}(y \cdot x^2) = x^3$$

Integrating both sides with respect to $x$:

$$\int \frac{d}{dx}(y \cdot x^2) dx = \int x^3 dx$$

This yields the general solution:

$$y \cdot x^2 = \frac{x^4}{4} + C$$

### Pinpointing the Curve

We have a family of curves, but we need the specific one that passes through $(1, 1)$. We substitute $x = 1$ and $y = 1$ into our general solution:

$$1 \cdot (1)^2 = \frac{1^4}{4} + C \Rightarrow 1 = \frac{1}{4} + C$$

Solving for the constant, we find $C = \frac{3}{4}$.

Plugging this back into our general solution and dividing by $x^2$, we obtain the final result:

**$$y = \frac{x^2}{4} + \frac{3}{4x^2}$$**

Animated Solution for Mathematics - Differential Equations: The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is

Select Answer:

Visualized Solution

Analyze the Differential Equation

Convert to Standard Form

Identify $P (x)$ and $Q (x)$

Calculate Integrating Factor

Simplify Integrating Factor

Apply General Solution Formula

Integrate

Isolate $y$

Apply Initial Condition

Solve for $C$

Final Specific Solution

Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Cleaning Phase

The Magic of the Integrating Factor

The Execution

Pinpointing the Curve

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation xdxdy​+2y=x2(x=0) with y(1)=1, is