Question

Find the particular solution of the differential equation $\log \left( \dfrac{dy}{dx} \right)=3x+4y$ , given that y = 0 when x = 0.

Answer 1

Hint: To solve this question, we will first exponential on both side, then we will write the right hand side using property of exponential, ${{e}^{a+b}}={{e}^{a}}\cdot {{e}^{b}}$. We will get variables of x and y separated then we will separate variables with dx and dy respectively and then integrate. After that, we will substitute the initial value condition and hence, obtain the solution.

Complete step by step answer:
Before we solve this question, let us see what is the differential equation, what is the solution of the differential equation and what is the type of solution.
A differential equation is an equation which contains independent variable, dependent variable and derivatives of dependent variable. For example, $\dfrac{dy}{dx}+y=x$ , where f ( x ) = y ( x ).
Solution of a differential equation is an equation free of derivatives of a dependent variable, which satisfies it’s differential equation.
General solution is a solution of a differential equation which contains arbitrary constants equal to order of differential equation.
Particular solution is a solution obtained from a general solution, by assigning some values to arbitrary constants.
Now, here differential equation is $\log \left( \dfrac{dy}{dx} \right)=3x+4y$
We can re – write $\log \left( \dfrac{dy}{dx} \right)=3x+4y$ as
$\log \left( \dfrac{dy}{dx} \right)=3x+4y$
$\left( \dfrac{dy}{dx} \right)={{e}^{3x+4y}}$, as $\log a=x$ then $a={{e}^{x}}$ .
Using, property of exponential, ${{e}^{a+b}}={{e}^{a}}\cdot {{e}^{b}}$ ,
$\left( \dfrac{dy}{dx} \right)={{e}^{3x}}\cdot {{e}^{4y}}$
We know that the differential equation can be solved by the variable separable if we can take dx and function on one side and dy and function of y on the other side.
So, \[\left( \dfrac{1}{{{e}^{4y}}} \right)dy={{e}^{3x}}dx\cdot \]
We know, $\dfrac{1}{{{e}^{a}}}={{e}^{-a}}$ , we get
\[{{e}^{-4y}}dy={{e}^{3x}}dx\cdot \]
Integrating both sides,
\[\int{{{e}^{-4y}}dy}=\int{{{e}^{3x}}dx}\]
We know that, \[\int{{{e}^{ax}}dx=\dfrac{{{e}^{ax}}}{a}}\]
So, \[-\dfrac{{{e}^{-4y}}}{4}=\dfrac{{{e}^{3x}}}{3}+C\]
Now, in question it is given that, y ( 0 ) = 0
So, putting y = 0 and x = 0 in \[-\dfrac{{{e}^{-4y}}}{4}=\dfrac{{{e}^{3x}}}{3}+C\], we get
\[-\dfrac{{{e}^{-4(0)}}}{4}=\dfrac{{{e}^{3(0)}}}{3}+C\]
\[-\dfrac{1}{4}=\dfrac{1}{3}+C\]
\[C=-\dfrac{1}{4}-\dfrac{1}{3}\]
On solving, we get
\[C=-\dfrac{7}{12}\]
Putting, \[C=-\dfrac{7}{12}\] in \[-\dfrac{{{e}^{-4y}}}{4}=\dfrac{{{e}^{3x}}}{3}+C\], we get
\[-\dfrac{{{e}^{-4y}}}{4}=\dfrac{{{e}^{3x}}}{3}-\dfrac{7}{12}\]
Or, \[\dfrac{{{e}^{3x}}}{3}+\dfrac{{{e}^{-4y}}}{4}=\dfrac{7}{12}\]
Or, \[4{{e}^{3x}}+3{{e}^{-4y}}=7\]

Hence, solution of differential equation $\log \left( \dfrac{dy}{dx} \right)=3x+4y$is \[4{{e}^{3x}}+3{{e}^{-4y}}=7\].


Note: While solving differential equation, firstly re – arrange the differential equation in such a way that you can get a differential equation in some known form such as variable separable, homogeneous or linear differential equation. While finding a particular solution, finding the value of constant very accurately as a particular solution is a unique solution for a pair of values of x and y.