NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The document NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
All you need of JEE at this link: JEE

Question 81: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This equation is in the form of a linear differential equation as:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 82: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This equation is the form of a linear differential equation as:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev  

Substituting the value of NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required general solution of the given differential equation.

 

Question 83: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This equation is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 84: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This equation is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 85: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 86: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 87: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 88: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now,NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Therefore,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting C = –2 in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the required solution of the given differential equation is NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 89: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now, y = 0 at x = 1.

Therefore,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required general solution of the given differential equation.

 

Question 90: NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now,NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Therefore, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting C = 4 in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required particular solution of the given differential equation.

Question 91: Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (xy) is equal to the sum of the coordinates of the point.

ANSWER : - Let F (xy) be the curve passing through the origin.

At point (xy), the slope of the curve will beNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

According to the given information:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The curve passes through the origin.

Therefore, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the required equation of curve passing through the origin isNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 92: Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

ANSWER : - Let F (xy) be the curve and let (xy) be a point on the curve. The slope of the tangent to the curve at (xy) is NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

According to the given information:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general equation of the curve is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Therefore, equation (1) becomes:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The curve passes through point (0, 2).

Therefore, equation (2) becomes:

0 2 – 4 = Ce0

⇒ – 2 = C

⇒ C = – 2

Substituting C = –2 in equation (2), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required equation of the curve.

Question 93: The integrating factor of the differential equationNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev is

A. ex                                   

B. ey                                   

C. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev                                                   

D. x

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The integrating factor (I.F) is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the correct answer is C.

Question 94: The integrating factor of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev is

A. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev                                

B. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev                              

C. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev                                

D. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The integrating factor (I.F) is given by the relation,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the correct answer is D.

 

Question 95: For each of the differential equations given below, indicate its order and degree (if defined).

(i) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev    
(ii) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

(iii) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

ANSWER : - (i) The differential equation is given as:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The highest order derivative present in the differential equation isNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev. Thus, its order is two. The highest power raised to NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevis one. Hence, its degree is one.

(ii) The differential equation is given as:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The highest order derivative present in the differential equation isNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev. Thus, its order is one. The highest power raised to NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevis three. Hence, its degree is three.

(iii) The differential equation is given as:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The highest order derivative present in the differential equation isNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev. Thus, its order is four.

However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

Question 96: For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

(ii) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

(iii) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

(iv) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

ANSWER : - (i) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Again, differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now, on substituting the values of NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevand NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin the differential equation, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

⇒ L.H.S. ≠ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Again, differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now, on substituting the values of NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevand NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin the L.H.S. of the given differential equation, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the given function is a solution of the corresponding differential equation.

(iii) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Again, differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting the value of NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev in the L.H.S. of the given differential equation, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the given function is a solution of the corresponding differential equation.

(iv) NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating both sides with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting the value of NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin the L.H.S. of the given differential equation, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the given function is a solution of the corresponding differential equation.

Question 97: Form the differential equation representing the family of curves given by NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevwhere a is an arbitrary constant.

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

From equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

On substituting this value in equation (3), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the differential equation of the family of curves is given as NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 98: Prove that NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevis the general solution of differential equationNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev, where c is a parameter.

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a homogeneous equation. To simplify it, we need to make the substitution as:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting the values of y and NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting the values of I1 and I2 in equation (3), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Therefore, equation (2) becomes:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the given result is proved.

 

Question 99: Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

ANSWER : - The equation of a circle in the first quadrant with centre (aa) and radius (a) which touches the coordinate axes is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating equation (1) with respect to x, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting the value of a in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the required differential equation of the family of circles is NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 100: Find the general solution of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 101: Show that the general solution of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevis given by (x    1) = (1 – – y – 2xy), where is parameter

 

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the given result is proved.

Question 102: Find the equation of the curve passing through the point NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevwhose differential equation is, NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The differential equation of the given curve is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The curve passes through point NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

On substituting NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the required equation of the curve is NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

 

Question 103: Find the particular solution of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev, given that y = 1 when x = 0

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting these values in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now, y = 1 at x = 0.

Therefore, equation (2) becomes:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin equation (2), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required particular solution of the given differential equation.

 

Question 104: Solve the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Differentiating it with respect to y, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

From equation (1) and equation (2), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 105: Find a particular solution of the differential equationNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev, given that = – 1, when x = 0 (Hint: put x – yt)

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting the values of x – and NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now, y = –1 at = 0.

Therefore, equation (3) becomes:

log 1 = 0 – 1 C

⇒ C = 1

Substituting C = 1 in equation (3) we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required particular solution of the given differential equation.

Question 106: Solve the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This equation is a linear differential equation of the form

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Question 107: Find a particular solution of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev, given that y = 0 when NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This equation is a linear differential equation of the form

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now,NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Therefore, equation (1) becomes:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevin equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required particular solution of the given differential equation.

Question 108: Find a particular solution of the differential equationNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev, given that y = 0 when x = 0

ANSWER : -

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting this value in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Now, at x = 0 and y = 0, equation (2) becomes:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Substituting C = 1 in equation (2), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is the required particular solution of the given differential equation.

Question 109: The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

ANSWER : - Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

log kt + C … (1)

In the year 1999, t = 0 and y = 20000.

Therefore, we get:

log 20000 = C … (2)

In the year 2004, t = 5 and = 25000.

Therefore, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

In the year 2009, t = 10 years.

Now, on substituting the values of tk, and C in equation (1), we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the population of the village in 2009 will be 31250.

Question 110: The general solution of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevis

A. xy = C                    

B. = Cy2                        

C. = Cx                   

D. y = Cx2

 

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Integrating both sides, we get:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the correct answer is C.

Question 111: The general solution of a differential equation of the type NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRevis

A. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev   
B. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

C. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev 
D. NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

ANSWER : - The integrating factor of the given differential equationNCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the differential equation is given by,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the correct answer is C.

Question 112: The general solution of the differential equation NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev is

A. xeyx2 = C
B. xey + y2 = C
C. yexx2 = C
Dyey + x2 = C

ANSWER : - The given differential equation is:

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

This is a linear differential equation of the form

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The general solution of the given differential equation is given by,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

Hence, the correct answer is C.

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

ppt

,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

,

Sample Paper

,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

,

shortcuts and tricks

,

Important questions

,

Previous Year Questions with Solutions

,

Summary

,

Extra Questions

,

NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

,

Semester Notes

,

Exam

,

video lectures

,

mock tests for examination

,

Viva Questions

,

pdf

,

study material

,

past year papers

,

Free

,

practice quizzes

,

MCQs

,

Objective type Questions

;