Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.5

Question 1. If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by \( M \frac { dV }{ dt } = F – kV \), where k is a constant. Express V in terms of t given that \( V = 0 \) when \( t = 0 \).
Answer:
Given the differential equation: \( M \frac { dV }{ dt } = F – kV \)
We can rewrite this equation to separate the variables:
\( \frac { dV }{ F – kV } = \frac { dt }{ M } \)
Now, we integrate both sides of the equation:
\( \int \frac { dV }{ F – kV } = \int \frac { dt }{ M } \)
To integrate the left side, we use a substitution or the standard integral form of \( \frac { 1 }{ ax+b } \).
\( \frac { \log (F – kV) }{ -k } = \frac { t }{ M } + \log C \)
This can be rearranged as:
\( \log (F – kV) = \frac { -kt }{ M } - k \log C \)
Let's combine the constant terms. Since C is an arbitrary constant, \( -k \log C \) can be absorbed into a new constant, or we can use \( \log C \) on the other side.
\( \log (F – kV) - \log C = \frac { -kt }{ M } \)
Using logarithm properties, \( \log A - \log B = \log (\frac { A }{ B }) \):
\( \log \left( \frac { F – kV }{ C } \right) = \frac { -kt }{ M } \)
To remove the logarithm, we take the exponential of both sides:
\( \frac { F – kV }{ C } = e^{\frac { -kt }{ M }} \)
Rearranging this, we get:
\( F – kV = C e^{\frac { -kt }{ M }} \) ... (1)
Now, we use the initial condition: when \( t = 0 \), \( V = 0 \). Substitute these values into equation (1):
\( F – k(0) = C e^{\frac { -k(0) }{ M }} \)
\( F – 0 = C e^0 \)
\( F = C \times 1 \)
\( C = F \)
Substitute the value of C back into equation (1):
\( F – kV = F e^{\frac { -kt }{ M }} \)
To express V in terms of t, we rearrange the equation:
\( kV = F - F e^{\frac { -kt }{ M }} \)
\( kV = F (1 - e^{\frac { -kt }{ M }}) \)
\( V = \frac { F }{ k } (1 - e^{\frac { -kt }{ M }}) \)
This equation shows how the velocity changes over time, starting from rest.
In simple words: We are given an equation that describes how a car's speed changes. We need to find a formula for the speed (V) at any time (t). We first separate the variables and integrate both sides. Then, we use the starting condition (speed is zero when time is zero) to find the constant in our solution. Finally, we rearrange the formula to show speed (V) by itself.

🎯 Exam Tip: Remember to clearly state the initial conditions and substitute them correctly to find the integration constant. This is a crucial step for full marks in such problems.

Question 2. The velocity v, of a parachute falling vertically satisfies the equation \( v\frac { dv }{ dx } = g(1 - \frac { v^2 }{ k^2 }) \) where g and k are constants. If v and x are both initially zero, find v in terms of x.
Answer:
Given the differential equation:
\( v\frac { dv }{ dx } = g\left(1 - \frac { v^2 }{ k^2 }\right) \)
First, simplify the right side of the equation:
\( v\frac { dv }{ dx } = g\left(\frac { k^2 - v^2 }{ k^2 }\right) \)
Now, we separate the variables, putting all v terms on one side and x terms on the other:
\( \frac { v }{ k^2 - v^2 } dv = \frac { g }{ k^2 } dx \)
To make the integration easier, we multiply both sides by -2:
\( \frac { -2v }{ k^2 - v^2 } dv = \frac { -2g }{ k^2 } dx \)
Next, we integrate both sides of the equation:
\( \int \frac { -2v }{ k^2 - v^2 } dv = \int \frac { -2g }{ k^2 } dx \)
The left side is of the form \( \int \frac { f'(v) }{ f(v) } dv = \log |f(v)| \). Here, if \( f(v) = k^2 - v^2 \), then \( f'(v) = -2v \).
So, integrating gives:
\( \log (k^2 - v^2) = \frac { -2gx }{ k^2 } + \log C \)
We rearrange the terms to solve for C:
\( \log (k^2 - v^2) - \log C = \frac { -2gx }{ k^2 } \)
Using logarithm properties, \( \log A - \log B = \log (\frac { A }{ B }) \):
\( \log \left( \frac { k^2 - v^2 }{ C } \right) = \frac { -2gx }{ k^2 } \)
Take the exponential of both sides to remove the logarithm:
\( \frac { k^2 - v^2 }{ C } = e^{\frac { -2gx }{ k^2 }} \)
So, \( k^2 - v^2 = C e^{\frac { -2gx }{ k^2 }} \) ... (1)
Now, apply the initial condition: when \( x = 0 \), \( v = 0 \). Substitute these into equation (1):
\( k^2 - (0)^2 = C e^{\frac { -2g(0) }{ k^2 }} \)
\( k^2 = C e^0 \)
\( k^2 = C \times 1 \)
\( C = k^2 \)
Substitute the value of C back into equation (1):
\( k^2 - v^2 = k^2 e^{\frac { -2gx }{ k^2 }} \)
To find v in terms of x, rearrange the equation:
\( v^2 = k^2 - k^2 e^{\frac { -2gx }{ k^2 }} \)
Factor out \( k^2 \):
\( v^2 = k^2 \left[ 1 - e^{\frac { -2gx }{ k^2 }} \right] \)
This equation describes the square of the velocity, which shows how the speed changes with distance fallen. This particular equation is often used to model motion with air resistance, where terminal velocity can be observed.
In simple words: We start with a given equation that links a parachute's speed (v) to its height (x). We rearrange the equation to put similar parts together, then integrate both sides. Using the starting point where both speed and height are zero, we find a missing constant. Finally, we put everything together to get a formula for the speed (v) based on the height (x).

🎯 Exam Tip: When dealing with differential equations, correctly separating variables is the first critical step. Pay close attention to the sign changes during integration and when substituting initial conditions.

Question 3. Find the equation of the curve whose slope is \( \frac { y-1 }{ x^2+x } \) and which passes through the point \( (1, 0) \).
Answer:
The slope of a curve is given by \( \frac { dy }{ dx } \). We are given:
\( \frac { dy }{ dx } = \frac { y-1 }{ x^2+x } \)
First, we separate the variables. Bring all y terms to one side and all x terms to the other:
\( \frac { dy }{ y-1 } = \frac { dx }{ x^2+x } \)
To integrate the right side, we need to simplify \( \frac { 1 }{ x^2+x } \). We can factor the denominator as \( x(x+1) \).
So, \( \frac { 1 }{ x(x+1) } \) can be split into partial fractions:
\( \frac { 1 }{ x(x+1) } = \frac { A }{ x } + \frac { B }{ x+1 } \)
Multiplying both sides by \( x(x+1) \), we get:
\( 1 = A(x+1) + B(x) \)
To find A, set \( x = 0 \):
\( 1 = A(0+1) + B(0) \)
\( 1 = A \)
To find B, set \( x = -1 \):
\( 1 = A(-1+1) + B(-1) \)
\( 1 = 0 - B \)
\( B = -1 \)
So, \( \frac { 1 }{ x^2+x } = \frac { 1 }{ x } - \frac { 1 }{ x+1 } \)
Now substitute this back into our separated differential equation:
\( \frac { dy }{ y-1 } = \left( \frac { 1 }{ x } - \frac { 1 }{ x+1 } \right) dx \)
Integrate both sides:
\( \int \frac { dy }{ y-1 } = \int \left( \frac { 1 }{ x } - \frac { 1 }{ x+1 } \right) dx \)
\( \log |y-1| = \log |x| - \log |x+1| + \log |C| \)
Using logarithm properties, \( \log A - \log B = \log (\frac { A }{ B }) \) and \( \log A + \log B = \log (AB) \):
\( \log |y-1| = \log \left| \frac { x }{ x+1 } \right| + \log |C| \)
\( \log |y-1| = \log \left| \frac { Cx }{ x+1 } \right| \)
Taking the exponential of both sides:
\( y-1 = \frac { Cx }{ x+1 } \)
This is the general equation of the curve. Now, we use the given point \( (1, 0) \) to find the value of C.
Substitute \( x = 1 \) and \( y = 0 \):
\( 0-1 = \frac { C(1) }{ 1+1 } \)
\( -1 = \frac { C }{ 2 } \)
\( C = -2 \)
Substitute C back into the equation of the curve:
\( y-1 = \frac { -2x }{ x+1 } \)
To find y in terms of x, add 1 to both sides:
\( y = 1 - \frac { 2x }{ x+1 } \)
To combine the terms on the right side, find a common denominator:
\( y = \frac { (x+1) - 2x }{ x+1 } \)
\( y = \frac { x+1 - 2x }{ x+1 } \)
\( y = \frac { 1-x }{ x+1 } \)
This is the specific equation of the curve that passes through the point \( (1,0) \). The process of separating variables and using partial fractions is common in solving these types of differential equations.
In simple words: We are given the slope of a curve and a point it goes through. We write the slope as a differential equation, then split the x and y terms apart. We use a trick called partial fractions to integrate the x part. After integrating, we use the given point to find the constant number (C) in our equation. Finally, we put it all together to get the exact equation for the curve.

🎯 Exam Tip: Always remember to use partial fraction decomposition when the denominator is a product of linear factors, as it simplifies the integration process significantly. Also, don't forget to substitute the given point to find the specific solution.

Question 4. Solve the following differential equations:
(i) \( \frac { dy }{ dx } = \frac { \sqrt { 1-y^2 } }{ \sqrt { 1-x^2 } } \)
(ii) \( ydx + (1 + x^2) \tan^{-1} x dy = 0 \)
(iii) \( \sin \left( \frac { dy }{ dx } \right) = a, y (0) = 1 \)
(iv) \( \frac { dy }{ dx } = e^{x+y} + x^3 e^{-y} \)
(v) \( (e^y + 1) \cos x dx + e^y \sin x dy = 0 \)
(vi) \( (ydx - xdy) \cot \left( \frac { x }{ y } \right) = ny^2 dx \)
(vii) \( \frac { dy }{ dx } - x\sqrt { 25-x^2 } = 0 \)
(viii) \( x \cos y dy = e^x (x \log x + 1) dx \)
(ix) \( \tan y \frac { dy }{ dx } = \cos (x + y) + \cos (x - y) \)
(x) \( \frac { dy }{ dx } = \tan^2 (x + y) \)
Answer:
(i) Given: \( \frac { dy }{ dx } = \frac { \sqrt { 1-y^2 } }{ \sqrt { 1-x^2 } } \)
Separate the variables:
\( \frac { dy }{ \sqrt { 1-y^2 } } = \frac { dx }{ \sqrt { 1-x^2 } } \)
Integrate both sides:
\( \int \frac { dy }{ \sqrt { 1-y^2 } } = \int \frac { dx }{ \sqrt { 1-x^2 } } \)
Using the standard integral \( \int \frac { dz }{ \sqrt { 1-z^2 } } = \sin^{-1} z \):
\( \sin^{-1} y = \sin^{-1} x + C \)

(ii) Given: \( ydx + (1 + x^2) \tan^{-1} x dy = 0 \)
Separate the variables:
\( ydx = - (1 + x^2) \tan^{-1} x dy \)
\( \frac { dx }{ (1+x^2) \tan^{-1} x } = - \frac { dy }{ y } \)
Let \( t = \tan^{-1} x \). Then \( dt = \frac { 1 }{ 1+x^2 } dx \).
Substitute this into the equation:
\( \frac { dt }{ t } = - \frac { dy }{ y } \)
Integrate both sides:
\( \int \frac { dt }{ t } = - \int \frac { dy }{ y } \)
\( \log |t| = - \log |y| + \log |C| \)
\( \log |t| = \log \left| \frac { C }{ y } \right| \)
Substitute back \( t = \tan^{-1} x \):
\( \log | \tan^{-1} x | = \log \left| \frac { C }{ y } \right| \)
Taking the exponential of both sides:
\( \tan^{-1} x = \frac { C }{ y } \)
\( y \tan^{-1} x = C \)

(iii) Given: \( \sin \left( \frac { dy }{ dx } \right) = a, y (0) = 1 \)
From the equation, we can write:
\( \frac { dy }{ dx } = \sin^{-1} (a) \)
Separate the variables and integrate:
\( \int dy = \int \sin^{-1} (a) dx \)
\( y = (\sin^{-1} a) x + C \) ... (1)
Now, apply the initial condition: \( y (0) = 1 \). Substitute \( x = 0 \) and \( y = 1 \) into (1):
\( 1 = (\sin^{-1} a) (0) + C \)
\( 1 = 0 + C \)
\( C = 1 \)
Substitute \( C = 1 \) back into equation (1):
\( y = (\sin^{-1} a) x + 1 \)
This can also be written as:
\( y - 1 = (\sin^{-1} a) x \)
\( \frac { y-1 }{ x } = \sin^{-1} a \)
\( \sin \left( \frac { y-1 }{ x } \right) = a \)

(iv) Given: \( \frac { dy }{ dx } = e^{x+y} + x^3 e^{-y} \)
We can rewrite \( e^{x+y} \) as \( e^x e^y \):
\( \frac { dy }{ dx } = e^x e^y + x^3 e^{-y} \)
Multiply the entire equation by \( e^y \) to simplify:
\( e^y \frac { dy }{ dx } = e^x e^{2y} + x^3 \)
This is not a simple separation. Let's try another approach: factor out \( e^{-y} \) on the right side.
\( \frac { dy }{ dx } = e^{-y} (e^{2y} e^x + x^3) \)
No, let's keep it simpler as given:
\( \frac { dy }{ dx } = e^x e^y + x^3 e^{-y} \)
Bring all \( e^y \) terms to the left:
\( \frac { dy }{ dx } - e^x e^y = x^3 e^{-y} \)
\( e^y \frac { dy }{ dx } - e^x e^{2y} = x^3 \) (This does not help)
Let's try to get all \( e^y \) with dy:
\( \frac { dy }{ dx } = e^y (e^x + x^3 e^{-2y}) \) (Still not right)
Let's re-examine the OCR for the original equation for (iv): `\frac { dy }{ dx }ex+y + X³, ey`. This is garbled. The solution starts with `\frac { dy }{ e^y } = dx(e^x + x^3)`. This means the original equation was likely of the form \( \frac { dy }{ dx } = e^y (e^x + x^3) \). Let's use this form.
Assume the question was: \( \frac { dy }{ dx } = e^x e^y + x^3 e^y \)
\( \frac { dy }{ dx } = e^y (e^x + x^3) \)
Separate variables:
\( \frac { dy }{ e^y } = (e^x + x^3) dx \)
Integrate both sides:
\( \int e^{-y} dy = \int (e^x + x^3) dx \)
\( -e^{-y} = e^x + \frac { x^4 }{ 4 } + C \)
Rearrange the terms to make the constant positive, or combine them:
\( e^x + e^{-y} + \frac { x^4 }{ 4 } + C = 0 \)
We can write the constant C with a different sign if preferred, so the final form is:
\( e^x + e^{-y} + \frac { x^4 }{ 4 } = C \)

(v) Given: \( (e^y + 1) \cos x dx + e^y \sin x dy = 0 \)
Separate the variables. Divide by \( (e^y + 1) \sin x \):
\( \frac { \cos x }{ \sin x } dx + \frac { e^y }{ e^y + 1 } dy = 0 \)
This can be written as:
\( \cot x dx + \frac { e^y }{ e^y + 1 } dy = 0 \)
Integrate both sides:
\( \int \cot x dx + \int \frac { e^y }{ e^y + 1 } dy = \int 0 \)
The integral of \( \cot x \) is \( \log |\sin x| \).
For the second integral, let \( u = e^y + 1 \), then \( du = e^y dy \). So \( \int \frac { du }{ u } = \log |u| \).
\( \log |\sin x| + \log |e^y + 1| = \log |C| \)
Using logarithm properties, \( \log A + \log B = \log (AB) \):
\( \log | (e^y + 1) \sin x | = \log |C| \)
Taking the exponential of both sides:
\( (e^y + 1) \sin x = C \)

(vi) Given: \( (ydx - xdy) \cot \left( \frac { x }{ y } \right) = ny^2 dx \)
This is a homogeneous differential equation. Divide by \( y^2 \) on the left and by \( \cot(\frac{x}{y}) \) on the right:
\( \frac { ydx - xdy }{ y^2 } \cot \left( \frac { x }{ y } \right) = n dx \)
The term \( \frac { ydx - xdy }{ y^2 } \) is the differential of \( \frac { x }{ y } \). That is, \( d\left( \frac { x }{ y } \right) = \frac { ydx - xdy }{ y^2 } \).
Let \( t = \frac { x }{ y } \). Then \( dt = \frac { ydx - xdy }{ y^2 } \).
Substitute these into the equation:
\( dt \cot t = n dx \)
Separate variables:
\( \cot t dt = n dx \)
Integrate both sides:
\( \int \cot t dt = \int n dx \)
\( \log |\sin t| = nx + C \)
To remove the logarithm, take the exponential of both sides:
\( \sin t = e^{nx+C} \)
\( \sin t = e^{nx} e^C \)
Let \( e^C = A \) (another constant).
\( \sin t = A e^{nx} \)
Substitute back \( t = \frac { x }{ y } \):
\( \sin \left( \frac { x }{ y } \right) = A e^{nx} \)

(vii) Given: \( \frac { dy }{ dx } - x\sqrt { 25-x^2 } = 0 \)
Rearrange the equation to separate variables:
\( \frac { dy }{ dx } = x\sqrt { 25-x^2 } \)
\( dy = x\sqrt { 25-x^2 } dx \)
Integrate both sides:
\( \int dy = \int x\sqrt { 25-x^2 } dx \)
For the right side integral, use substitution. Let \( t = 25-x^2 \).
Then \( dt = -2x dx \), so \( x dx = - \frac { 1 }{ 2 } dt \).
The integral becomes:
\( y = \int \sqrt{t} \left( -\frac { 1 }{ 2 } \right) dt \)
\( y = - \frac { 1 }{ 2 } \int t^{1/2} dt \)
Use the power rule for integration \( \int z^n dz = \frac { z^{n+1} }{ n+1 } \):
\( y = - \frac { 1 }{ 2 } \left( \frac { t^{1/2+1} }{ 1/2+1 } \right) + C \)
\( y = - \frac { 1 }{ 2 } \left( \frac { t^{3/2} }{ 3/2 } \right) + C \)
\( y = - \frac { 1 }{ 2 } \times \frac { 2 }{ 3 } t^{3/2} + C \)
\( y = - \frac { 1 }{ 3 } t^{3/2} + C \)
Substitute back \( t = 25-x^2 \):
\( y = - \frac { 1 }{ 3 } (25-x^2)^{3/2} + C \)
We can also write this as:
\( 3y = - (25-x^2)^{3/2} + 3C \)
Let \( 3C = C_1 \) (another constant).
\( 3y + (25-x^2)^{3/2} = C_1 \)

(viii) Given: \( x \cos y dy = e^x (x \log x + 1) dx \)
Separate the variables. Divide both sides by x:
\( \cos y dy = e^x \left( \frac { x \log x + 1 }{ x } \right) dx \)
Simplify the term in the parenthesis:
\( \cos y dy = e^x \left( \frac { x \log x }{ x } + \frac { 1 }{ x } \right) dx \)
\( \cos y dy = e^x \left( \log x + \frac { 1 }{ x } \right) dx \)
Integrate both sides:
\( \int \cos y dy = \int e^x \left( \log x + \frac { 1 }{ x } \right) dx \)
The left side integral is \( \sin y \).
The right side integral is of the form \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).
Here, if \( f(x) = \log x \), then \( f'(x) = \frac { 1 }{ x } \). This matches the form.
So, \( \sin y = e^x \log x + C \)

(ix) Given: \( \tan y \frac { dy }{ dx } = \cos (x + y) + \cos (x - y) \)
First, simplify the right side using the trigonometric identity \( \cos(A+B) + \cos(A-B) = 2 \cos A \cos B \).
Here, \( A = x \) and \( B = y \).
So, \( \cos (x + y) + \cos (x - y) = 2 \cos x \cos y \)
The equation becomes:
\( \tan y \frac { dy }{ dx } = 2 \cos x \cos y \)
Separate the variables. Divide both sides by \( \cos y \):
\( \frac { \tan y }{ \cos y } dy = 2 \cos x dx \)
We know that \( \frac { \tan y }{ \cos y } = \tan y \sec y \).
So, \( \tan y \sec y dy = 2 \cos x dx \)
Integrate both sides:
\( \int \tan y \sec y dy = \int 2 \cos x dx \)
The integral of \( \tan y \sec y \) is \( \sec y \).
\( \sec y = 2 \sin x + C \)

(x) Given: \( \frac { dy }{ dx } = \tan^2 (x + y) \)
This is a non-linear differential equation. We use a substitution to simplify it.
Let \( t = x + y \).
Differentiate t with respect to x:
\( \frac { dt }{ dx } = 1 + \frac { dy }{ dx } \)
So, \( \frac { dy }{ dx } = \frac { dt }{ dx } - 1 \)
Substitute \( t \) and \( \frac { dy }{ dx } \) into the original equation:
\( \frac { dt }{ dx } - 1 = \tan^2 t \)
Rearrange to separate variables:
\( \frac { dt }{ dx } = 1 + \tan^2 t \)
We know the trigonometric identity \( 1 + \tan^2 t = \sec^2 t \).
So, \( \frac { dt }{ dx } = \sec^2 t \)
Separate the variables:
\( \frac { dt }{ \sec^2 t } = dx \)
We know that \( \frac { 1 }{ \sec^2 t } = \cos^2 t \).
So, \( \cos^2 t dt = dx \)
To integrate \( \cos^2 t \), use the identity \( \cos^2 t = \frac { 1 + \cos 2t }{ 2 } \):
\( \frac { 1 + \cos 2t }{ 2 } dt = dx \)
Integrate both sides:
\( \int \frac { 1 + \cos 2t }{ 2 } dt = \int dx \)
\( \frac { 1 }{ 2 } \int (1 + \cos 2t) dt = \int dx \)
\( \frac { 1 }{ 2 } \left( t + \frac { \sin 2t }{ 2 } \right) = x + C \)
Substitute back \( t = x + y \):
\( \frac { 1 }{ 2 } \left( (x+y) + \frac { \sin (2(x+y)) }{ 2 } \right) = x + C \)
This can also be written as \( (x+y) + \frac { 2 \sin(x+y) \cos(x+y) }{ 2 } = 2x + 2C \)
\( (x+y) + \sin(x+y)\cos(x+y) = 2x + C_1 \) (where \( C_1 = 2C \))
The key is to apply the correct trigonometric identities for integration.
In simple words: For each equation, we aim to put all parts with 'y' and 'dy' on one side and all parts with 'x' and 'dx' on the other. Then, we integrate both sides using standard formulas. Sometimes, we need to use a substitution or trigonometric identity to make the integration possible. The goal is to find a formula for 'y' or 'v' in terms of 'x' or 't'.

🎯 Exam Tip: For problems with multiple sub-parts, clearly label each solution. Pay attention to common integral forms, trigonometric identities, and substitution methods which are key to solving various differential equation types.