JEE Main 2020 (7 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Q: Why can't we just integrate both sides of $e^y(\frac{dy}{dx} - 1) = e^x$ directly?

That is a great question! The reason is that the variables $x$ and $y$ are 'entangled' inside the exponential function and the derivative term. In a separable differential equation, you need to be able to write the equation in the form $f(y)dy = g(x)dx$. Because of the term $e^{x-y}$, you cannot isolate $y$ and $x$ on opposite sides without first performing a substitution to break that entanglement.

Q: What happens to the constant $C$ after we integrate?

The constant $C$ is essential! It represents the family of all possible solutions. To find the specific solution for this problem, we use the initial condition $y(0) = 0$. By plugging $x=0$ and $y=0$ into our general solution, we can solve for the exact value of $C$. Without this step, we would only have a general solution, not the particular one required to find $y(1)$.

Select Answer:

Visualized Solution

Analyze the Given Equation

Given differential equation: $e^{y} (\frac{d y}{d x} - 1) = e^{x}$
Initial condition: $y (0) = 0$
Goal: Find $y (1)$

Isolate the Derivative $\frac{d y}{d x}$

Divide both sides by $e^{y}$ : $\frac{d y}{d x} - 1 = \frac{e ^{x}}{e ^{y}}$
Simplify using exponent laws: $\frac{d y}{d x} - 1 = e^{x - y}$
Isolate $\frac{d y}{d x}$ : $\frac{d y}{d x} = e^{x - y} + 1$

Identify the Substitution $x - y = t$

Notice the term $x - y$ in the exponent.
Let $x - y = t$
This substitution will transform the equation into a separable form.

Differentiate the Substitution

Differentiate $t = x - y$ with respect to $x$ .
$\frac{d t}{d x} = 1 - \frac{d y}{d x}$
Rearrange to express $\frac{d y}{d x}$ in terms of $t$ :
$\frac{d y}{d x} = 1 - \frac{d t}{d x}$

Substitute back into the Equation

Original simplified equation: $\frac{d y}{d x} = e^{x - y} + 1$
Substitute $\frac{d y}{d x} = 1 - \frac{d t}{d x}$ and $x - y = t$ :
$1 - \frac{d t}{d x} = e^{t} + 1$

Simplify and Separate Variables

Subtract $1$ from both sides: $- \frac{d t}{d x} = e^{t}$
Rearrange to separate variables $t$ and $x$ :
$\frac{d t}{e ^{t}} = - d x$
$- e^{- t} d t = d x$

Integrate Both Sides

Integrate both sides: $\int - e^{- t} d t = \int d x$
The integral of $- e^{- t}$ is $e^{- t}$ .
The integral of $d x$ is $x$ .
Result: $e^{- t} = x + C$

Back-Substitute $t = x - y$

Recall our initial substitution: $t = x - y$
Substitute $t$ back into the integrated equation:
$e^{- (x - y)} = x + C$
Simplify the exponent: $e^{y - x} = x + C$

Apply Initial Condition $y (0) = 0$

Use the given condition: when $x = 0$ , $y = 0$ .
Substitute these values into the general solution:
$e^{0 - 0} = 0 + C$
$e^{0} = C ⟹ 1 = C$
Particular solution: $e^{y - x} = x + 1$

Find $y (1)$

We need to find the value of $y$ when $x = 1$ .
Substitute $x = 1$ into the particular solution:
$e^{y - 1} = 1 + 1 ⟹ e^{y - 1} = 2$
Take the natural logarithm ( $ln$ ) on both sides:
$y - 1 = ln (2)$
$y = 1 + ln (2)$

Final Conclusion

Final Answer: $y (1) = 1 + lo g_{e} 2$
The correct option is (4).
Key Takeaway: Recognizing the substitution $x - y = t$ was the critical step to transform a non-separable equation into a separable one.

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The Sigma Insight: Variable Separable Method

Analyzing the Setup

We are given the differential equation:

e^{y} (\frac{d y}{d x} - 1) = e^{x}

with the initial condition

y (0) = 0

. Our objective is to determine the value of

y (1)

The Elegant Transformation

First, we isolate the derivative by dividing both sides by

e^{y}

\frac{d y}{d x} - 1 = \frac{e ^{x}}{e ^{y}} = e^{x - y}

Rearranging the terms, we obtain:

\frac{d y}{d x} = e^{x - y} + 1

To simplify this, we introduce the substitution

t = x - y

. Differentiating with respect to

x

yields:

\frac{d t}{d x} = 1 - \frac{d y}{d x} ⟹ \frac{d y}{d x} = 1 - \frac{d t}{d x}

Substituting these into our differential equation, we get:

1 - \frac{d t}{d x} = e^{t} + 1

The Beauty of Cancellation

Subtracting

1

from both sides of the equation simplifies the expression significantly:

- \frac{d t}{d x} = e^{t}

This is a separable differential equation. We can rewrite it as:

- e^{- t} d t = d x

Integrating both sides, we find:

\int - e^{- t} d t = \int d x ⟹ e^{- t} = x + C

The Final Reveal

Substituting

t = x - y

back into the equation, we have:

e^{- (x - y)} = x + C ⟹ e^{y - x} = x + C

Applying the initial condition

y (0) = 0

e^{0 - 0} = 0 + C ⟹ C = 1

The particular solution is therefore

e^{y - x} = x + 1

. To find

y (1)

, we set

x = 1

e^{y - 1} = 1 + 1 = 2

Taking the natural logarithm of both sides:

y - 1 = ln (2)

Thus, the final result is:

$y = 1 + ln (2)$

Common Traps & FAQs

Why can't we just integrate both sides of

e^{y} (\frac{d y}{d x} - 1) = e^{x}

directly?

That is a great question! The reason is that the variables

x

and

y

are 'entangled' inside the exponential function and the derivative term. In a separable differential equation, you need to be able to write the equation in the form

f (y) d y = g (x) d x

. Because of the term

e^{x - y}

, you cannot isolate

y

and

x

on opposite sides without first performing a substitution to break that entanglement.

How did you know to choose

t = x - y

as the substitution?

In differential equations, we often look for patterns. Notice that the exponent is

x - y

and the derivative term is

(\frac{d y}{d x} - 1)

. If you differentiate

x - y

, you get

1 - \frac{d y}{d x}

, which is exactly the negative of the term in the parentheses. This 'matching' of the derivative of the substitution to the existing terms in the equation is the hallmark of a perfect substitution.

What happens to the constant

C

after we integrate?

The constant

C

is essential! It represents the family of all possible solutions. To find the specific solution for this problem, we use the initial condition

y (0) = 0

. By plugging

x = 0

and

y = 0

into our general solution, we can solve for the exact value of

C

. Without this step, we would only have a general solution, not the particular one required to find

y (1)

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Select Answer:

Visualized Solution

Analyze the Given Equation

Isolate the Derivative $\frac{d y}{d x}$

Identify the Substitution $x - y = t$

Differentiate the Substitution

Substitute back into the Equation

Simplify and Separate Variables

Integrate Both Sides

Back-Substitute $t = x - y$

Apply Initial Condition $y (0) = 0$

Find $y (1)$

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Elegant Transformation

The Beauty of Cancellation

The Final Reveal

Similar Questions

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

If $y = y (x)$ is the solution of the differential equation $\frac{5 + e ^{x}}{2 + y} \frac{d y}{d x} + e^{x} = 0$ satisfying $y (0) = 1$ then a value of $y (lo g_{e} 13)$ is :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $e^{x} 1 - y^{2} d x + (\frac{y}{x}) d y = 0, y (1) = - 1$ . Then the value of $(y (3))^{2}$ is equal to:

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

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

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, ey(dxdy​−1)=ex such that y(0)=0, then y(1) is equal to

Select Answer:

Visualized Solution

Analyze the Given Equation

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

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

Differentiate the Substitution

Substitute back into the Equation

Simplify and Separate Variables

Integrate Both Sides

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

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

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

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Elegant Transformation

The Beauty of Cancellation

The Final Reveal

Similar Questions

.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 (1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0, then 6[y′(0)+(y(loge​3​))2] is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2y2x+y−2x​,y(0)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x=x(y) be the solution of the differential equation 2yex/y2dx+(y2−4xex/y2)dy=0 such that x(1)=0. Then, x(e) is equal to

.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 2+y5+ex​dxdy​+ex=0 satisfying y(0)=1 then a value of y(loge​13) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+2x−12x−y(2y−1)​=0,x,y>0,y(1)=1, then y(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation ex1−y2​dx+(xy​)dy=0,y(1)=−1. Then the value of (y(3))2 is equal to:

.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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dy=eαx+ydx;α∈N. If y(loge​2)=loge​2 and y(0)=loge​(1/2), then the value of α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (1+y2)etanxdx+cos2x(1+e2tanx)dy=0,y(0)=1. Then y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution curve y=y(x) of the differential equation (1+y2)(1+loge​x)dx+xdy=0,x>0 passes through the point (1,1) and y(e)=β+tan(3/2)α−tan(3/2)​, then α+2β is

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Isolate the Derivative $\frac{d y}{d x}$

Identify the Substitution $x - y = t$

Back-Substitute $t = x - y$

Apply Initial Condition $y (0) = 0$

Find $y (1)$

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

If $y = y (x)$ is the solution of the differential equation $\frac{5 + e ^{x}}{2 + y} \frac{d y}{d x} + e^{x} = 0$ satisfying $y (0) = 1$ then a value of $y (lo g_{e} 13)$ is :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $e^{x} 1 - y^{2} d x + (\frac{y}{x}) d y = 0, y (1) = - 1$ . Then the value of $(y (3))^{2}$ is equal to:

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

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is