
If \[\dfrac{{dy}}{{dx}} = {e^{ - 2y}}\] and \[y = 0\] for \[x = 5\]. Then find the value of \[x\] for \[y = 3\].
A. \[{e^5}\]
B. \[{e^6} + 1\]
C. \[\dfrac{{{e^9} + 1}}{2}\]
D. \[\log_{e}6 \]
Answer
162.9k+ views
Hint: Here, the first order differential equation is given. First, simplify the given equation by rearranging the terms. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve the integrals by using the U-substitution method on the left-hand side and get the solution of the differential equation. Then, substitute the given values in the solution and find the value of the integration constant. In the end, substitute the values and integration constant in the solution and solve it to get the required answer.
Formula Used: \[\int {ndx} = x + c\] , where \[n\] is a number and \[c\] is an integration constant.
\[\int {{e^x}dx} = {e^x} + c\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = {e^{ - 2y}}\] and \[y = 0\] for \[x = 5\].
Simplify the given equation.
\[\dfrac{{dy}}{{{e^{ - 2y}}}} = dx\]
\[ \Rightarrow {e^{2y}}dy = dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {{e^{2y}}dy} = \int {dx} \] \[.....\left( 1 \right)\]
It is difficult to find the integral of the above equation.
So, apply the substitution method on the left-hand side.
Substitute \[2y = u\].
Differentiating the substitute equation, we get
\[2dy = du\]
\[ \Rightarrow dy = \dfrac{{du}}{2}\]
Then, we get the equation \[\left( 1 \right)\] as
\[\int {\dfrac{{{e^u}}}{2}du} = \int {dx} \]
Take the constant term outside of the integral.
\[\dfrac{1}{2}\int {{e^u}du} = \int {dx} \]
Now solve the integrals by using the formulas \[\int {{e^x}dx} = {e^x} + c\], and \[\int {ndx} = x + c\].
We get,
\[\dfrac{1}{2}{e^u} = x + c\]
Resubstitute the value of \[u\].
\[\dfrac{1}{2}{e^{2y}} = x + c\]
The solution of the given differential equation is \[\dfrac{1}{2}{e^{2y}} = x + c\].
Now to find the value of the integration constant \[c\], substitute \[y = 0\] and \[x = 5\] in the solution.
\[\dfrac{1}{2}{e^{2\left( 0 \right)}} = 5 + c\]
\[ \Rightarrow \dfrac{1}{2}{e^0} = 5 + c\]
\[ \Rightarrow \dfrac{1}{2}\left( 1 \right) = 5 + c\]
\[ \Rightarrow c = \dfrac{1}{2} - 5\]
\[ \Rightarrow c = - \dfrac{9}{2}\]
We have to find the value of \[x\] for \[y = 3\].
So, substitute \[y = 3\], and \[c = - \dfrac{9}{2}\] in the equation \[\left( 1 \right)\].
Then,
\[\dfrac{1}{2}{e^{2\left( 3 \right)}} = x - \dfrac{9}{2}\]
Solve the above equation.
\[\dfrac{1}{2}{e^6} = \dfrac{{2x - 9}}{2}\]
\[ \Rightarrow {e^6} = 2x - 9\]
\[ \Rightarrow {e^6} + 9 = 2x\]
Divide both sides by 2.
\[x = \dfrac{{{e^6} + 9}}{2}\]
Option ‘C’ is correct
Note: Students often forget to calculate the value of the integration constant while solving this type of questions.
Always remember to calculate the value of the integration constant by using the given values or information.
Formula Used: \[\int {ndx} = x + c\] , where \[n\] is a number and \[c\] is an integration constant.
\[\int {{e^x}dx} = {e^x} + c\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = {e^{ - 2y}}\] and \[y = 0\] for \[x = 5\].
Simplify the given equation.
\[\dfrac{{dy}}{{{e^{ - 2y}}}} = dx\]
\[ \Rightarrow {e^{2y}}dy = dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {{e^{2y}}dy} = \int {dx} \] \[.....\left( 1 \right)\]
It is difficult to find the integral of the above equation.
So, apply the substitution method on the left-hand side.
Substitute \[2y = u\].
Differentiating the substitute equation, we get
\[2dy = du\]
\[ \Rightarrow dy = \dfrac{{du}}{2}\]
Then, we get the equation \[\left( 1 \right)\] as
\[\int {\dfrac{{{e^u}}}{2}du} = \int {dx} \]
Take the constant term outside of the integral.
\[\dfrac{1}{2}\int {{e^u}du} = \int {dx} \]
Now solve the integrals by using the formulas \[\int {{e^x}dx} = {e^x} + c\], and \[\int {ndx} = x + c\].
We get,
\[\dfrac{1}{2}{e^u} = x + c\]
Resubstitute the value of \[u\].
\[\dfrac{1}{2}{e^{2y}} = x + c\]
The solution of the given differential equation is \[\dfrac{1}{2}{e^{2y}} = x + c\].
Now to find the value of the integration constant \[c\], substitute \[y = 0\] and \[x = 5\] in the solution.
\[\dfrac{1}{2}{e^{2\left( 0 \right)}} = 5 + c\]
\[ \Rightarrow \dfrac{1}{2}{e^0} = 5 + c\]
\[ \Rightarrow \dfrac{1}{2}\left( 1 \right) = 5 + c\]
\[ \Rightarrow c = \dfrac{1}{2} - 5\]
\[ \Rightarrow c = - \dfrac{9}{2}\]
We have to find the value of \[x\] for \[y = 3\].
So, substitute \[y = 3\], and \[c = - \dfrac{9}{2}\] in the equation \[\left( 1 \right)\].
Then,
\[\dfrac{1}{2}{e^{2\left( 3 \right)}} = x - \dfrac{9}{2}\]
Solve the above equation.
\[\dfrac{1}{2}{e^6} = \dfrac{{2x - 9}}{2}\]
\[ \Rightarrow {e^6} = 2x - 9\]
\[ \Rightarrow {e^6} + 9 = 2x\]
Divide both sides by 2.
\[x = \dfrac{{{e^6} + 9}}{2}\]
Option ‘C’ is correct
Note: Students often forget to calculate the value of the integration constant while solving this type of questions.
Always remember to calculate the value of the integration constant by using the given values or information.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
