The solution of the differential equation dy/dx = (x-y)², when y(1) = 1, is: (A) log_e |(2-x)/(2-y)|

Q: Wait, why do we substitute $x-y=t$ instead of just trying to separate variables directly?

That is a fantastic question! If you look at the original equation $\frac{dy}{dx} = (x-y)^2$, the variables $x$ and $y$ are 'tangled' inside the square. There is no algebraic way to isolate $x$ on one side and $y$ on the other. By substituting $t = x-y$, we essentially perform a change of variables that 'untangles' the expression, turning a non-separable equation into a simple, beautiful variable-separable one.

Q: Why do we need the initial condition $y(1)=1$?

Great catch! When we integrate both sides of a differential equation, we always get a constant of integration, $C$. This $C$ represents an infinite family of curves. The initial condition $y(1)=1$ acts like a 'pin' that fixes our specific curve to a single point in the coordinate plane. It allows us to solve for the unique value of $C$ that satisfies the problem's specific constraints.

Q: I got $\frac{1}{2} \ln |\frac{1+t}{1-t}| = x + C$, but the options have a negative sign outside the log. Did I do something wrong?

Not at all! You are on the right track. Remember the logarithmic property $\ln(\frac{a}{b}) = -\ln(\frac{b}{a})$. If your derivation leads to a positive log, you can simply flip the fraction inside the absolute value to introduce a negative sign. This is a common technique in JEE problems to match the final answer with the given options.

Question

The solution of the differential equation $\frac{dy}{dx} = (x-y)^2$, when $y(1) = 1$, is:

Options:

(A) $\log_e |\frac{2-x}{2-y}| = x-y$

(B) $-\log_e |\frac{1-x+y}{1+x-y}| = 2(x-1)$

(C) $-\log_e |\frac{1+x-y}{1-x+y}| = x+y-2$

(D) $\log_e |\frac{2-y}{2-x}| = 2(y-1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex differential equation:

$$\frac{dy}{dx} = (x-y)^2$$

At first glance, it looks intimidating. The variables $x$ and $y$ are locked in a tight embrace inside that square.

You might be tempted to expand it, but that only makes the mess worse. In the world of JEE Advanced, when you see a linear combination like $(x-y)$ trapped inside a function, you are looking at an invitation to use a powerful tool: **Substitution**.

### The Power of Substitution

We start by defining a new variable, $t = x-y$. This is the spark. By letting $t = x-y$, we are essentially simplifying the geometry of the problem.

Now, we need to see how this affects the derivative. Differentiating both sides with respect to $x$, we get:

$$\frac{d}{dx}(x) - \frac{d}{dx}(y) = \frac{dt}{dx}$$

This simplifies to $1 - \frac{dy}{dx} = \frac{dt}{dx}$. Rearranging this, we find that:

$$\frac{dy}{dx} = 1 - \frac{dt}{dx}$$

Now, watch the magic happen. We substitute this back into our original equation:

$$1 - \frac{dt}{dx} = t^2$$

Suddenly, the $y$ has vanished! We are left with a clean, elegant equation in terms of $t$ and $x$:

$$\frac{dt}{dx} = 1 - t^2$$

### The Elegance of Separation

With the equation in the form $\frac{dt}{dx} = 1 - t^2$, we can easily separate the variables. By moving the terms, we get:

$$\frac{dt}{1-t^2} = dx$$

This is the heart of the solution. We have transformed a tangled mess into a standard integral. We know from our calculus toolkit that:

$$\int \frac{dt}{a^2-t^2} = \frac{1}{2a} \ln \left| \frac{a+t}{a-t} \right| + C$$

Applying this with $a=1$, we get:

$$\frac{1}{2} \ln \left| \frac{1+t}{1-t} \right| = x + C$$

### Finding the Unique Path

We are almost there, but we must return to our original variables. Substituting $t = x-y$ back into our equation, we get:

$$\frac{1}{2} \ln \left| \frac{1+(x-y)}{1-(x-y)} \right| = x + C$$

This simplifies to:

$$\frac{1}{2} \ln \left| \frac{1+x-y}{1-x+y} \right| = x + C$$

Now, we use the initial condition $y(1)=1$. When $x=1$, $y=1$. Plugging these in, we get:

$$\frac{1}{2} \ln \left| \frac{1+1-1}{1-1+1} \right| = 1 + C$$

Since $\ln(1) = 0$, we find that $0 = 1 + C$, which means **$C = -1$**.

### The Final Celebration

Substituting $C = -1$ back into our equation, we have:

$$\frac{1}{2} \ln \left| \frac{1+x-y}{1-x+y} \right| = x - 1$$

Multiplying by $2$, we get:

$$\ln \left| \frac{1+x-y}{1-x+y} \right| = 2(x-1)$$

To match the standard forms, we use the property $\ln(\frac{a}{b}) = -\ln(\frac{b}{a})$ to flip the fraction, giving us:

$$-\ln \left| \frac{1-x+y}{1+x-y} \right| = 2(x-1)$$

This is the exact form we were looking for! You have successfully navigated the complexity and arrived at the solution. Keep this substitution trick in your arsenal—it is a classic JEE favorite.

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

Select Answer:

Visualized Solution

Analyzing $\frac{d y}{d x} = (x - y)^{2}$

The Substitution $x - y = t$

Differentiating $x - y = t$

Expressing $\frac{d y}{d x}$

Separating Variables $t$ and $x$

Setting up $\int \frac{d t}{1 - t ^{2}}$

Executing the Integration

Back Substitution of $t$

Applying Initial Condition $y (1) = 1$

Finding the Constant $C$

Final Simplification

The Way Forward

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Substitution

The Elegance of Separation

Finding the Unique Path

The Final Celebration

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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 dxdy​=(x−y)2, when y(1)=1, is:

Select Answer:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​=(x−y)2

The Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−y=t

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−y=t

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Separating Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫1−t2dt​

Executing the Integration

Back Substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Applying Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(1)=1

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C