**RD Sharma Solutions for Class 12 Maths Exercise 22.10 Chapter 22 Differential Equations** are provided here. Our specialists will teach you many methods to tackle questions in the exam and manage time in the examination. Students can refer to RD Sharma Solutions for Class 12 Maths Chapter 22 and practise various questions daily to perform well in the final exams.

## RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 10

### Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 10

EXERCISE 22.10

**Question. 1**

**Solution:**

From the question, it is given that,

(dy/dx) + 2y = e^{3x} … [equation (i)]

The given linear differential equation is compared with (dy/dx) + py = Q

So, p = 2, Q = e^{3x}

IF = e^{∫pdx}

= e^{∫2dx}

= e^{2x}

Then, multiplying both sides of equation (i) by IF,

e^{2x}(dy/dx) + e^{2x} 2y = e^{2x} × e^{3x}

e^{2x}(dy/dx) + e^{2x} 2y = e^{5x} … [because a^{m} × a^{n} = a^{m + n}]

Now, integrating the above equation with respect to x,

ye^{2x} = ∫e^{5x} dx + c

ye^{2x} = (e^{5x}/5) + c

y = (e^{5x}/5) + ce^{-2x}

**Question. 2**

**Solution:**

From the question, it is given that,

4(dy/dx) + 8y = 5e^{-3x}

Dividing both sides by 4, we get,

dy/dx + 2y = 5e^{-3x}/4 … [equation (i)]

The given linear differential equation is compared with, (dy/dx) + py = Q

So, p = 2, Q = 5e^{-3x}/4

IF = e^{∫pdx}

= e^{∫2dx}

= e^{2x}

Then, multiplying both sides of equation (i) by IF,

e^{2x}(dy/dx) + e^{2x} 2y = e^{2x} × 5e^{-3x}/4

e^{2x}(dy/dx) + e^{2x} 2y = 5e^{-x}/4

Now, integrating the above equation with respect to x,

ye^{2x} = ∫5e^{-x}/4 dx+ c

ye^{2x} = (-5/4) e^{-x} + c

y = (-5/4) e^{-3x} + ce^{-2x}

**Question. 3**

**Solution:**

From the question, it is given that,

(dy/dx) + 2y = 6e^{x} … [equation (i)]

The given linear differential equation is compared with (dy/dx) + py = Q

So, p = 2, Q = 6e^{x}

IF = e^{∫pdx}

= e^{∫2dx}

= e^{2x}

Then, multiplying both sides of equation (i) by IF,

e^{2x}(dy/dx) + e^{2x} 2y = e^{2x} × 6e^{x}

e^{2x}(dy/dx) + e^{2x} 2y = 6e^{3x} … [because a^{m} × a^{n} = a^{m + n}]

Now, integrating the above equation with respect to x,

ye^{2x} = ∫6e^{3x} dx + c

ye^{2x} = (6/3) e^{3x} + c

y = 2e^{3x} + ce^{-2x}

**Question. 4**

**Solution:**

From the question, it is given that,

(dy/dx) + y = e^{-2x} … [equation (i)]

The given linear differential equation is compared with, (dy/dx) + py = Q

So, p = 1, Q = e^{-2x}

IF = e^{∫pdx}

= e^{∫1dx}

= e^{x}

Then, multiplying both sides of equation (i) by IF,

e^{x}(dy/dx) + e^{x} y = e^{x} × e^{-2x}

e^{x}(dy/dx) + e^{x} y = e^{-x} … [because a^{m} × a^{n} = a^{m + n}]

Now, integrating the above equation with respect to x,

ye^{x} = ∫e^{-x} dx + c

ye^{x} = e^{-x}/-1 + c

y = -e^{-2x} + ce^{-x}

**Question. 5**

**Solution:**

From the question, it is given that,

x(dy/dx) = x + y

dy/dx = (x + y)/x

dy/dx = 1 + (y/x) … [equation (i)]

The given linear differential equation is compared with (dy/dx) + py = Q

So, p = – 1/x, Q = 1

IF = e^{∫pdx}

= e^{∫-1/xdx}

= e^{-log x}

= e^{log(1/x)}

= 1/x

Then, multiplying both sides of equation (i) by IF,

Y(1/x) = ∫1 (1/x) dx + c

ye^{x} = log x + c

y = x log x + cx

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