diff --git a/solutions/CMakeLists.txt b/solutions/CMakeLists.txt
index 09cff2a..68396c4 100644
--- a/solutions/CMakeLists.txt
+++ b/solutions/CMakeLists.txt
@@ -1,3 +1,5 @@
 include_directories(${CMAKE_CURRENT_SOURCE_DIR})
 
 add_subdirectory(chapter-01)
+add_subdirectory(chapter-02)
+add_subdirectory(chapter-03)
diff --git a/solutions/chapter-02/CMakeLists.txt b/solutions/chapter-02/CMakeLists.txt
new file mode 100644
index 0000000..ca5acc6
--- /dev/null
+++ b/solutions/chapter-02/CMakeLists.txt
@@ -0,0 +1,12 @@
+include_directories(${CMAKE_CURRENT_SOURCE_DIR})
+
+# The next lines represent a list of all the executables in the folder (chapter)
+# There is typically one executable per exercise.
+# As you solve exercises, you need to add them here.
+# The syntax is add_executable(executable_name source_file_name)
+
+add_executable(ch-02-ex-02-solution ch-02-ex-02-solution.cpp)
+add_executable(ch-02-ex-03-solution ch-02-ex-03-solution.cpp)
+add_executable(ch-02-ex-04-solution ch-02-ex-04-solution.cpp)
+add_executable(ch-02-ex-05-solution ch-02-ex-05-solution.cpp)
+add_executable(ch-02-ex-06-solution ch-02-ex-06-solution.cpp)
\ No newline at end of file
diff --git a/solutions/chapter-02/ch-02-ex-02-solution.cpp b/solutions/chapter-02/ch-02-ex-02-solution.cpp
new file mode 100644
index 0000000..62a3f60
--- /dev/null
+++ b/solutions/chapter-02/ch-02-ex-02-solution.cpp
@@ -0,0 +1,70 @@
+/*
+ * chapter-02-exercise-02.cpp
+ *
+ * Using the "if" clause
+ *
+ * Created on: Aug 19, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  double p, q, x, y;
+  int j;
+
+  /*
+   * Read p
+   */
+  std::cout << "Please input value of p (double): ";
+  std::cin >> p;
+  /*
+   * Read q
+   */
+  std::cout << "Please input value of q (double): ";
+  std::cin >> q;
+  /*
+   * Read y
+   */
+  std::cout << "Please input value of y (double): ";
+  std::cin >> y;
+  /*
+   * Read j
+   */
+  std::cout << "Please input value of j (integer): ";
+  std::cin >> j;
+
+  /*
+   * Set x to 5 if either p is greater than or equal to q, or the variable
+   * j is not equal to 10
+   */
+  if ((p >= q) || (j != 10)) {
+    x = 5;
+  }
+
+  /*
+   * Set x to 5 if both y is greater than or equal to q, and the variable j
+   * is equal to 20. Otherwise set x to p
+   */
+  if ((y >= q) && (j == 20)) {
+    x = 5;
+  } else {
+    x = p;
+  }
+
+  /*
+   * Set x according to rule at ex 2.2.3
+   */
+  if (p > q) {
+    x = 0;
+  } else if (j == 10) {
+    x = 1;
+  } else {
+    x = 2;
+  }
+
+  // Print out x
+  std::cout << "The value of x is: " << x << std::endl;
+
+  return 0;
+}
diff --git a/solutions/chapter-02/ch-02-ex-03-solution.cpp b/solutions/chapter-02/ch-02-ex-03-solution.cpp
new file mode 100644
index 0000000..9757792
--- /dev/null
+++ b/solutions/chapter-02/ch-02-ex-03-solution.cpp
@@ -0,0 +1,57 @@
+/*
+ * chapter-02-exercise-04.cpp
+ *
+ * Using the while loop
+ *
+ * Created on: Aug 19, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  int number = 0, sum = 0;
+
+  /*
+   * Point 1. Terminate reading numbers when -1 is input by the user
+   */
+  while (number >= 0) {
+    sum += number;
+    std::cout << "Please input numbers. Terminate with -1 :";
+    std::cin >> number;
+  }
+  std::cout << "The sum of the numbers is: " << sum << std::endl;
+
+  /*
+   * Point 2. Terminate reading numbers when -1 is input by the user
+   * or when sum exceedes 100.
+   */
+  do {
+    std::cout << "Please input numbers. Terminate with -1 :";
+    std::cin >> number;
+    if (number != -1)
+      sum += number;
+  } while ((number >= 0) && (sum <= 100));
+  std::cout << "The sum of the numbers is: " << sum << std::endl;
+
+  /*
+   * Point 3. Reset sum by inputting -2
+   */
+  do {
+    std::cout << "Please input numbers. Terminate with -1. Reset with -2 :";
+    std::cin >> number;
+    switch (number) {
+    case -1:
+      break;
+    case -2:
+      sum = 0;
+      break;
+    default:
+      sum += number;
+      break;
+    }
+  } while ((number >= 0) && (sum <= 100));
+  std::cout << "The sum of the numbers is: " << sum << std::endl;
+
+  return 0;
+}
diff --git a/solutions/chapter-02/ch-02-ex-04-solution.cpp b/solutions/chapter-02/ch-02-ex-04-solution.cpp
new file mode 100644
index 0000000..0b0e09a
--- /dev/null
+++ b/solutions/chapter-02/ch-02-ex-04-solution.cpp
@@ -0,0 +1,112 @@
+/*
+ * chapter-02-exercise-04.cpp
+ *
+ * Vector operations with loops
+ *
+ * Created on: Aug 19, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  double u[3] = {1.0, 2.0, 3.0};
+  double v[3] = {6.0, 5.0, 4.0};
+  double A[3][3] = {{1.0, 5.0, 0.0}, {7.0, 1.0, 2.0}, {0.0, 0.0, 1.0}};
+  double B[3][3] = {{-2.0, 0.0, 1.0}, {1.0, 0.0, 0.0}, {4.0, 1.0, 0.0}};
+  double w[3];
+  for (int i = 0; i < 3; ++i) {
+    w[i] = u[i] - 3.0 * v[i];
+  }
+
+  // x = u - v
+  double x[3] = {0.0, 0.0, 0.0};
+  for (int i = 0; i < 3; ++i) {
+    x[i] = u[i] - v[i];
+  }
+
+  // y = Au
+  double y[3] = {0.0, 0.0, 0.0};
+  for (int i = 0; i < 3; ++i) {
+    for (int j = 0; j < 3; ++j) {
+      y[i] += A[i][j] * u[j];
+    }
+  }
+
+  // z = Au - v
+  double z[3] = {0.0, 0.0, 0.0};
+  for (int i = 0; i < 3; ++i) {
+    z[i] -= v[i];
+    for (int j = 0; j < 3; ++j) {
+      z[i] += A[i][j] * u[j];
+    }
+  }
+
+  // C = 4A - 3B
+  double C[3][3];
+  for (int i = 0; i < 3; ++i) {
+    for (int j = 0; j < 3; ++j) {
+      C[i][j] = 4 * A[i][j] - 3 * B[i][j];
+    }
+  }
+
+  // D = AB
+  double D[3][3] = {{0.0, 0.0, 0.0}, {0.0, 0.0, 0.0}, {0.0, 0.0, 0.0}};
+  for (int i = 0; i < 3; ++i) {
+    for (int j = 0; j < 3; ++j) {
+      for (int k = 0; k < 3; ++k) {
+        D[i][j] += A[i][k] * B[k][j];
+      }
+    }
+  }
+
+  // Print w
+  std::cout << "w is: ";
+  for (int i = 0; i < 3; ++i) {
+    std::cout << w[i] << " ";
+  }
+  std::cout << std::endl;
+
+  // Print x
+  std::cout << "x is: ";
+  for (int i = 0; i < 3; ++i) {
+    std::cout << x[i] << " ";
+  }
+  std::cout << std::endl;
+
+  // Print y
+  std::cout << "y is: ";
+  for (int i = 0; i < 3; ++i) {
+    std::cout << y[i] << " ";
+  }
+  std::cout << std::endl;
+
+  // Print z
+  std::cout << "z is: ";
+  for (int i = 0; i < 3; ++i) {
+    std::cout << z[i] << " ";
+  }
+  std::cout << std::endl;
+
+  // Print C
+  std::cout << "C is:" << std::endl;
+  for (int i = 0; i < 3; ++i) {
+    for (int j = 0; j < 3; ++j) {
+      std::cout << C[i][j] << " ";
+    }
+    std::cout << std::endl;
+  }
+  std::cout << std::endl;
+
+  // Print D
+  std::cout << "D is:" << std::endl;
+  for (int i = 0; i < 3; ++i) {
+    for (int j = 0; j < 3; ++j) {
+      std::cout << D[i][j] << " ";
+    }
+    std::cout << std::endl;
+  }
+  std::cout << std::endl;
+
+  return 0;
+}
diff --git a/solutions/chapter-02/ch-02-ex-05-solution.cpp b/solutions/chapter-02/ch-02-ex-05-solution.cpp
new file mode 100644
index 0000000..61495a8
--- /dev/null
+++ b/solutions/chapter-02/ch-02-ex-05-solution.cpp
@@ -0,0 +1,37 @@
+/*
+ * chapter-02-exercise-05.cpp
+ *
+ * Inverse for 2x2 matrix
+ *
+ * Created on: Aug 19, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <cassert>
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  double A[2][2] = {{4, 10}, {1, 1}};
+
+  // Compute the determinant
+  double det = A[0][0] * A[1][1] - A[0][1] * A[1][0];
+  assert(det != 0);
+
+  // Compute the inverse
+  double Ainv[2][2];
+  Ainv[0][0] = A[1][1] / det;
+  Ainv[0][1] = -A[0][1] / det;
+  Ainv[1][0] = -A[1][0] / det;
+  Ainv[1][1] = A[0][0] / det;
+
+  // Print out inverse
+  std::cout << "The inverse of A is:" << std::endl;
+  for (int i = 0; i < 2; ++i) {
+    for (int j = 0; j < 2; ++j) {
+      std::cout << Ainv[i][j] << " ";
+    }
+    std::cout << std::endl;
+  }
+
+  return 0;
+}
diff --git a/solutions/chapter-02/ch-02-ex-06-solution.cpp b/solutions/chapter-02/ch-02-ex-06-solution.cpp
new file mode 100644
index 0000000..ed08f70
--- /dev/null
+++ b/solutions/chapter-02/ch-02-ex-06-solution.cpp
@@ -0,0 +1,38 @@
+/*
+ * chapter-02-exercise-05.cpp
+ *
+ * Newton-Raphson iteration for f(x) = exp(x) + x^3 + 5 = 0; x0 = 0;
+ *
+ * Created on: Aug 19, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <cmath>
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  // Define a stopping precision
+  double epsilon = 1e-6;
+  int max_iter = 1000;
+  double x_prev = 0.0;
+  double x_next = 0.0;
+
+  int num_iter = 0;
+  do {
+    x_prev = x_next;
+    x_next = x_prev - (exp(x_prev) + pow(x_prev, 3) + 5) /
+                          (exp(x_prev) + 3 * pow(x_prev, 2));
+    ++num_iter;
+    std::cout << "x[" << num_iter << "] = " << x_next << std::endl;
+  } while ((std::abs(x_next - x_prev) > epsilon) && (num_iter < max_iter));
+
+  if ((num_iter == max_iter) && (std::abs(x_next - x_prev) > epsilon)) {
+    std::cout << "Max number of iterations reached without convergence"
+              << std::endl;
+  } else {
+    std::cout << "The solution to exp(x) + x^3 + 5 = 0 is: " << x_next
+              << std::endl;
+  }
+
+  return 0;
+}
diff --git a/solutions/chapter-03/CMakeLists.txt b/solutions/chapter-03/CMakeLists.txt
new file mode 100644
index 0000000..1aa79f2
--- /dev/null
+++ b/solutions/chapter-03/CMakeLists.txt
@@ -0,0 +1,10 @@
+include_directories(${CMAKE_CURRENT_SOURCE_DIR})
+
+# The next lines represent a list of all the executables in the folder (chapter)
+# There is typically one executable per exercise.
+# As you solve exercises, you need to add them here.
+# The syntax is add_executable(executable_name source_file_name)
+
+add_executable(ch-03-ex-01-solution ch-03-ex-01-solution.cpp)
+add_executable(ch-03-ex-02-solution ch-03-ex-02-solution.cpp)
+add_executable(ch-03-ex-03-solution ch-03-ex-03-solution.cpp)
diff --git a/solutions/chapter-03/ch-03-ex-01-solution.cpp b/solutions/chapter-03/ch-03-ex-01-solution.cpp
new file mode 100644
index 0000000..bf46c7b
--- /dev/null
+++ b/solutions/chapter-03/ch-03-ex-01-solution.cpp
@@ -0,0 +1,47 @@
+/*
+ * chapter-03-exercise-01.cpp
+ *
+ * Print to file with a given precision
+ *
+ * Created on: Sep 2, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <fstream>
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  double x[4] = {0.0, 1.0, 1.0, 0.0};
+  double y[4] = {0.0, 0.0, 1.0, 1.0};
+
+  std::ofstream output_file("x_and_y.dat");
+  // Check that file is properly openened.
+  if (!output_file.is_open()) {
+    std::cout << "Error opening output file." << std::endl;
+    return 1;
+  }
+
+  // Output in scientific format with 10 significant figures
+  // and always showing sign
+  output_file.setf(std::ios::scientific);
+  output_file.setf(std::ios::showpos);
+  output_file.precision(10);
+  // Write x vector ending with a new line
+  for (int i = 0; i < 4; ++i) {
+    output_file << x[i] << " ";
+  }
+  output_file << "\n";
+  // Flush after writing vector x
+  output_file.flush();
+  // Do the same for y vector
+  for (int i = 0; i < 4; ++i) {
+    output_file << y[i] << " ";
+  }
+  output_file << "\n";
+  // Flush after writing vector y
+  output_file.flush();
+  // Close the file when finished
+  output_file.close();
+
+  return 0;
+}
diff --git a/solutions/chapter-03/ch-03-ex-02-solution.cpp b/solutions/chapter-03/ch-03-ex-02-solution.cpp
new file mode 100644
index 0000000..bb9b5d2
--- /dev/null
+++ b/solutions/chapter-03/ch-03-ex-02-solution.cpp
@@ -0,0 +1,48 @@
+/*
+ * chapter-03-exercise-02.cpp
+ *
+ * Count number of rows in a file
+ *
+ * Created on: Sep 2, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <fstream>
+#include <iostream>
+#include <string>
+
+int main(int argc, char *argv[]) {
+  std::ifstream read_file("x_and_y.dat");
+  if (!read_file.is_open()) {
+    std::cout << "Error opening file." << std::endl;
+    return 1;
+  }
+
+  int number_of_rows = 0;
+
+  // This loop will read an empty line between the final EOL and the EOF
+  // characters
+  while (!read_file.eof()) {
+    double dummy1, dummy2, dummy3, dummy4;
+    read_file >> dummy1 >> dummy2 >> dummy3 >> dummy4;
+    number_of_rows++;
+  }
+  std::cout << "Number of rows = " << number_of_rows << " (false)" << std::endl;
+
+  // We try again
+  number_of_rows = 0;
+  read_file.clear();
+  read_file.seekg(std::ios::beg);
+  while (!read_file.eof()) {
+    std::string line;
+    getline(read_file, line);
+    if (line.compare("") != 0) {
+      number_of_rows++;
+    }
+  }
+  std::cout << "Number of rows = " << number_of_rows << " (true)" << std::endl;
+
+  read_file.close();
+
+  return 0;
+}
diff --git a/solutions/chapter-03/ch-03-ex-03-solution.cpp b/solutions/chapter-03/ch-03-ex-03-solution.cpp
new file mode 100644
index 0000000..56daaf8
--- /dev/null
+++ b/solutions/chapter-03/ch-03-ex-03-solution.cpp
@@ -0,0 +1,64 @@
+/*
+ * chapter-03-exercise-03.cpp
+ *
+ * Implements the backward Euler method to solve:
+ * 		dy / dx = -y, y_0 = 1, 0 <= x <= 1
+ *
+ * 	The discretization for N grid points is:
+ * 		- step size: h = 1 / (N - 1)
+ * 		- grid points: x_n = n * h
+ * 		- solution: y_(n) - y_(n-1) / h = - y_n ; n = 1,2,...N-1
+ *
+ *
+ * Created on: Sep 2, 2012
+ * 	   Author: Radu Popescu <radu.popescu@epfl.ch>
+ */
+
+#include <cassert>
+#include <fstream>
+#include <iostream>
+
+int main(int argc, char *argv[]) {
+  // Read from the command line the number of grid points to use
+  int N;
+  std::cout << "Please input the number of grid points. N = ";
+  std::cin >> N;
+  assert(N > 1);
+
+  // Calculate step size h
+  double h = 1.0 / (N - 1);
+  std::cout << "Computed step size. h = " << h << std::endl;
+
+  // Start computing the grid points and the solution in these points
+  // At each step, we output the solution to disk
+  std::ofstream output_file("xy.dat");
+  if (!output_file.is_open()) {
+    std::cout << "Error opening output file." << std::endl;
+    return 1;
+  }
+  output_file.setf(std::ios::scientific);
+  output_file.setf(std::ios::showpos);
+  output_file.precision(9);
+
+  // x_(n-1) = x_0 = 0
+  double x_prev = 0;
+  // x_(n) = x_1 = h;
+  double x_next = h;
+  // y_(n-1) = y_0 = 1;
+  double y_prev = 1;
+  double y_next;
+
+  output_file << x_prev << " " << y_prev << std::endl;
+
+  while (x_next <= 1.0) {
+    y_next = y_prev / (h + 1.0);
+    output_file << x_next << " " << y_next << std::endl;
+    y_prev = y_next;
+    x_prev = x_next;
+    x_next = x_prev + h;
+  }
+
+  output_file.close();
+
+  return 0;
+}