If y = y(x) is the solution of the differential equation x dy/dx + 2y = x² satisfying y(1) = 1, then

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

Great question! We divide by $x$ because the standard form for a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$. Without this step, the equation is not in a form where the Integrating Factor method is applicable. Think of it as cleaning the workspace before you start the real construction.

Q: Why does $e^{2 \ln x}$ simplify to $x^2$? I always get confused with log properties.

That is a common point of confusion! Remember the property $a \ln b = \ln(b^a)$. So, $2 \ln x$ becomes $\ln(x^2)$. Since the exponential function $e^u$ and the natural logarithm $\ln u$ are inverse functions, $e^{\ln(x^2)}$ simply simplifies to $x^2$. It is one of the most elegant simplifications in calculus!

Q: Do we really need the constant $C$? Can't we just find the general solution?

The general solution gives us a whole family of curves. But the problem gives us a specific condition: $y(1) = 1$. This is our 'anchor'. Without finding $C$, we are just looking at a general shape; with $C$, we are looking at the exact path the particle or function takes. It is the difference between a general idea and a specific, concrete reality.

Question

If $y = y(x)$ is the solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ satisfying $y(1) = 1$, then $y(1/2)$ is equal to :

Options:

(A) $\frac{7}{64}$

(B) $\frac{13}{16}$

(C) $\frac{49}{16}$

(D) $\frac{1}{4}$

Sigma Percentile · Accepted Answer

### The Beauty of Linear Differential Equations

Welcome, future engineers! Today, we are going to embark on a journey through a classic problem that perfectly illustrates the elegance of linear differential equations.

Imagine you are standing before a mathematical landscape, and you are given a rule that describes how a function changes:

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

This isn't just a collection of symbols; it is a story of growth and change. Our goal is to find the specific path this function takes, anchored by the condition $y(1) = 1$, and then predict its value at $x = \frac{1}{2}$.

### Phase 1: The Clean Up

In the world of differential equations, we love order. The standard form for a linear differential equation is:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

Looking at our given equation, $x \frac{dy}{dx} + 2y = x^2$, we see that the coefficient of $\frac{dy}{dx}$ is $x$. This is a slight obstacle, but a simple one to overcome.

We divide the entire equation by $x$, transforming it into:

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

Now, the equation is in its pristine, standard form. We can clearly see that $P(x) = \frac{2}{x}$ and $Q(x) = x$. This clarity is the first step toward victory.

### Phase 2: The Magic Multiplier

Now, we introduce the most powerful tool in our arsenal: the Integrating Factor ($I.F.$). The $I.F.$ is designed to turn the left side of our equation into the derivative of a product, making it integrable.

The formula is:

$$I.F. = e^{\int P(x) dx}$$

Substituting our $P(x)$, we get:

$$I.F. = e^{\int \frac{2}{x} dx}$$

The integral of $\frac{2}{x}$ is $2 \ln x$. Using the beautiful properties of logarithms, $2 \ln x$ becomes $\ln(x^2)$.

Since $e$ and $\ln$ are inverse functions, they cancel out, leaving us with an incredibly simple $I.F. = x^2$. It is moments like these where the complexity of the math just melts away into something elegant.

### Phase 3: The Integration

With our $I.F.$ in hand, the general solution is given by:

$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$

Plugging in our values, we get:

$$y \cdot x^2 = \int x \cdot x^2 dx + C$$

This simplifies to:

$$y \cdot x^2 = \int x^3 dx + C$$

Integrating $x^3$ is straightforward using the power rule: $\frac{x^4}{4}$. So, our general solution is:

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

We have successfully navigated the integration!

### Phase 4: The Anchor

We are almost there. We have a family of curves, but we need the one that passes through $(1, 1)$.

By substituting $x = 1$ and $y = 1$ into our general solution, we get:

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

This simplifies to $1 = \frac{1}{4} + C$. Solving for $C$, we find $C = \frac{3}{4}$.

Now, our particular solution is locked in:

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

### Phase 5: The Final Victory

Finally, we need to find $y(\frac{1}{2})$. We substitute $x = \frac{1}{2}$ into our particular solution:

$$y \cdot \left(\frac{1}{2}\right)^2 = \frac{(\frac{1}{2})^4}{4} + \frac{3}{4}$$

This becomes:

$$y \cdot \frac{1}{4} = \frac{1/16}{4} + \frac{3}{4} \Rightarrow y \cdot \frac{1}{4} = \frac{1}{64} + \frac{3}{4}$$

Combining the fractions, we get:

$$y \cdot \frac{1}{4} = \frac{1 + 48}{64} = \frac{49}{64}$$

Multiplying both sides by $4$, we arrive at our final answer:

**$$y = \frac{49}{16}$$**

We have successfully traced the path of the function and found our target value. Well done!

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :

Select Answer:

Visualized Solution

The Given Equation

Standard Form Conversion

Identifying $P (x)$ and $Q (x)$

The Integrating Factor Setup

Calculating the Integrating Factor

The General Solution Formula

Integrating the RHS

Applying the Boundary Condition

Finding the Constant $C$

The Particular Solution

Evaluating at $x = \frac{1}{2}$

Final Calculation

The Sigma Insight: Linear Differential Equations

The Beauty of Linear Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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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; } If y=y(x) is the solution of the differential equation xdxdy​+2y=x2 satisfying y(1)=1, then y(1/2) is equal to :