Consider the differential equation, y² dx + (x - 1/y) dy = 0. If value of y is 1 when x = 1, then th

Q: Wait, why do we reject the standard $dy/dx$ approach? Can't we just solve it that way?

That is a fantastic question! While you *could* technically try to solve for $dy/dx$, you would end up with an equation where $y$ is non-linear and the terms are coupled in a way that makes separation of variables or standard linear methods impossible. By shifting your perspective to treat $x$ as the dependent variable and $y$ as the independent variable, the equation becomes linear, which is a much more elegant and solvable path.

Q: How do I know if a differential equation is linear in $x$ or $y$?

Look at the degree of the variables and their derivatives. If you see a term like $x$ (to the power of 1) and its derivative $dx/dy$ (also to the power of 1) without any products like $x \cdot dx/dy$, it is a strong candidate for being linear in $x$. Always check if you can rearrange the terms to fit the standard form $dx/dy + P(y)x = Q(y)$.

Q: Why did we use $t = -1/y$ for the integral? It looks confusing.

I understand it looks intimidating, but substitution is your best friend in calculus! We chose $t = -1/y$ because its derivative, $dt = (1/y^2) dy$, perfectly matches a part of our integral. By breaking $1/y^3$ into $(1/y) \cdot (1/y^2)$, we can substitute $t$ for $-1/y$ and $dt$ for $(1/y^2) dy$, turning a complex exponential integral into a simple $\int -t e^t dt$ which is easily solved by parts.

Question

Consider the differential equation, $y^2 dx + (x - \frac{1}{y}) dy = 0$. If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Options:

(A) $\frac{1}{2} + \frac{1}{\sqrt{e}}$

(B) $\frac{3}{2} - \sqrt{e}$

(C) $\frac{5}{2} + \frac{1}{\sqrt{e}}$

(D) $\frac{3}{2} - \frac{1}{\sqrt{e}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:
$$y^2 dx + \left(x - \frac{1}{y}\right) dy = 0$$

At first glance, this equation appears complex. However, attempting to force it into the standard $\frac{dy}{dx}$ form leads to a non-linear expression. The key to solving this problem is to shift our perspective and treat $x$ as the dependent variable and $y$ as the independent variable.

### The Strategic Rearrangement

We begin by moving the $dy$ term to the right side:
$$y^2 dx = \left(\frac{1}{y} - x\right) dy$$

Next, we divide the entire equation by $y^2 dy$ to isolate the derivative:
$$\frac{dx}{dy} = \frac{1}{y^3} - \frac{x}{y^2}$$

By rearranging the terms to bring the $x$ component to the left, we obtain the standard linear form:
$$\frac{dx}{dy} + \left(\frac{1}{y^2}\right)x = \frac{1}{y^3}$$

### The Magic of the Integrating Factor

The equation is now in the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$. We calculate the Integrating Factor ($IF$) as follows:
$$IF = e^{\int P(y) dy} = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$$

Multiplying the differential equation by this $IF$ allows us to express the left side as the derivative of a product:
$$\frac{d}{dy} \left(x e^{-\frac{1}{y}}\right) = \frac{1}{y^3} e^{-\frac{1}{y}}$$

### The Integral Journey

To find the general solution, we integrate both sides with respect to $y$:
$$x e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy$$

We use the substitution $t = -\frac{1}{y}$, which implies $dt = \frac{1}{y^2} dy$. Splitting the term $\frac{1}{y^3}$ into $\frac{1}{y} \cdot \frac{1}{y^2}$, the integral becomes:
$$\int (-t) e^t dt$$

Applying integration by parts, we solve $\int -t e^t dt = e^t(1 - t)$. Substituting back $t = -\frac{1}{y}$, we obtain:
$$x e^{-\frac{1}{y}} = e^{-\frac{1}{y}}\left(1 + \frac{1}{y}\right) + C$$

### Final Calculation

We apply the initial condition $x = 1$ when $y = 1$ to determine the constant $C$:
$$1 \cdot e^{-1} = e^{-1}(1 + 1) + C \implies C = -e^{-1}$$

Substituting $C$ back into the equation, we get the particular solution:
$$x = 1 + \frac{1}{y} - e^{\frac{1}{y}-1}$$

Finally, we evaluate $x$ at $y = 2$:
$$x = 1 + \frac{1}{2} - e^{\frac{1}{2}-1} = \frac{3}{2} - e^{-\frac{1}{2}}$$

The final answer is **$x = \frac{3}{2} - \frac{1}{\sqrt{e}}$**.

Animated Solution for Mathematics - Differential Equations: Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Converting to Standard Linear Form

Identifying $P (y)$ and $Q (y)$

Calculating the Integrating Factor ( $I F$ )

Setting up the General Solution

Substitution for Integration

Integration by Parts

Back Substitution

The General Solution Equation

Finding the Constant $C$

The Particular Solution

Final Calculation for $y = 2$

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Strategic Rearrangement

The Magic of the Integrating Factor

The Integral Journey

Final Calculation

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Select Answer:

Visualized Solution

Analyzing the Differential Equation

Converting to Standard Linear Form

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(y) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(y)

Calculating the Integrating Factor ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IF)