Question

The particular solution of differential equation \[\log \left( {\dfrac{{dy}}{{dx}}} \right) = 3x + 4y\] is, when \[y = 0 = x\]
A.\[4{e^{3x}} + 3{e^{ - 4y}} + 7 = 0\]
B.\[4{e^{3x}} - 3{e^{ - 4y}} - 7 = 0\]
C.\[4{e^{3x}} + 3{e^{ - 4y}} - 7 = 0\]
D.\[4{e^{3x}} - 3{e^{ - 4y}} + 7 = 0\]

Answer 1

Hint: Here we will first simplify the given differential equation using the properties of logarithmic functions. Then we will integrate the obtained equation and substitute the given values of the variable to find the constant. We will then back substitute the obtained constant term in the obtained equation to find the required solution.

Complete step-by-step answer:
We will first simplify the given differential equation using the properties of logarithmic functions.
The given differential equation is \[\log \left( {\dfrac{{dy}}{{dx}}} \right) = 3x + 4y\].
We know from the properties of the logarithmic function that
When \[\log a = x\] then \[a = {e^x}\]
So we will use the same property of logarithmic function in the above equation. Therefore, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{\left( {3x + 4y} \right)}}\]
We know the property of the exponential function that \[{e^{a + b}} = {e^a} \cdot {e^b}\].
So, we will use the same property of exponential function in the above equation. Therefore, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}} \cdot {e^{4y}}\]
On separating the variables, we get
\[ \Rightarrow \dfrac{{dy}}{{{e^{4y}}}} = {e^{3x}} \cdot dx\]
Now, we will integrate the terms on both sides of the equation.
\[ \Rightarrow \int {\dfrac{{dy}}{{{e^{4y}}}}} = \int {{e^{3x}} \cdot dx} \]
On further simplifying the terms, we get
\[ \Rightarrow \int {{e^{ - 4y}} \cdot dy} = \int {{e^{3x}} \cdot dx} \]
On integrating the terms, we get
\[ \Rightarrow \dfrac{{ - {e^{ - 4y}}}}{4} = \dfrac{{{e^{3x}}}}{3} + c\]
On further simplifying the terms, we get
\[\begin{array}{l} \Rightarrow 0 = \dfrac{{{e^{3x}}}}{3} + \dfrac{{{e^{ - 4y}}}}{4} + c\\ \Rightarrow \dfrac{{{e^{3x}}}}{3} + \dfrac{{{e^{ - 4y}}}}{4} + c = 0\end{array}\]
Taking LCM of the fractions, we get
\[ \Rightarrow \dfrac{{4{e^{3x}} + 3{e^{ - 4y}}}}{{12}} + c = 0\]
Multiplying both sides by 12, we get
\[ \Rightarrow 4{e^{3x}} + 3{e^{ - 4y}} + 12c = 0\] ……………. \[\left( 1 \right)\]
Now, we will substitute the given values of the variables \[x = 0\] and \[y = 0\]. Therefore, we get
\[ \Rightarrow 4{e^{3 \times 0}} + 3{e^{ - 4 \times 0}} + 12c = 0\]
Now, we will simplify the terms further.
\[ \Rightarrow 4{e^0} + 3{e^0} + 12c = 0\]
Now substituting \[{e^0} = 1\] in the above equation, we get
On adding the numbers, we get
\[ \Rightarrow 7 + 12c = 0\]
Now, subtracting 7 from both sides, we get
\[\begin{array}{l} \Rightarrow 7 + 12c - 7 = 0 - 7\\ \Rightarrow 12c = - 7\end{array}\]
Now, dividing both sides by the number 3, we get
\[ \Rightarrow \dfrac{{12c}}{{12}} = \dfrac{{ - 7}}{{12}}\]
\[ \Rightarrow c = \dfrac{{ - 7}}{{12}}\]
Now, we will substitute the value of the constant term \[c\] in equation \[\left( 1 \right)\]. Therefore, we get
\[4{e^{3x}} + 3{e^{ - 4y}} + 12 \times \left( {\dfrac{{ - 7}}{{12}}} \right) = 0\]
On multiplying the numbers, we get
\[ \Rightarrow 4{e^{3x}} + 3{e^{ - 4y}} - 7 = 0\]
Therefore, the required particular solution of the given differential equation is equal to \[4{e^{3x}} + 3{e^{ - 4y}} - 7 = 0\]
Hence, the correct option is option C.

Note: Here we have obtained the particular solution of the given differential equation and we have used various properties of the logarithmic function. The logarithmic function is defined as the function which is the inverse of the exponential function. Also, the value of the logarithm of the negative numbers is not defined. The logarithm of any positive number, whose base is a number, which is greater than zero and is not equal to the number one, is the power to which the base can be raised in order to obtain the given number.