Question

What is a solution to the differential equation \[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\] with \[y\left( 0 \right)=1\].

Answer 1

Hint: For the equation \[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\] will use VARIABLE SEPARABLE method to questions. With the help of this method, we can find the solution of the given differential equation in a simple and fast manner.
Also, we should know \[\int{{{e}^{x}}dx}={{e}^{x}}+c\] and \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c\](where \[c\] is constant) Before solving this question.

Complete step by step solution:
From the given question, we were given that
\[\Rightarrow \]\[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\] with \[y\left( 0 \right)=1\]
Consider equation \[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\]……………….(1)
Now we have to cross multiply the equation, we get
\[\Rightarrow \]\[y\times dy={{e}^{x}}\times dx\]……………(2)
As we separated all the \[y\]terms to one side and \[x\] to other side, we can use variable separable methods now.
Now apply integration on both sides. We get
\[\Rightarrow \]\[\int{y}\times dy=\int{{{e}^{x}}}\times dx\]……………..(3)
As we know from the basic integration formulas \[\int{{{e}^{x}}dx}={{e}^{x}}+c\] and \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c\](where \[c\] is constant) we can write \[\int{{{y}^{1}}dx=\dfrac{{{y}^{1+1}}}{1+1}}+{{c}_{1}}\] and \[\int{{{e}^{x}}}\times dx={{e}^{x}}+{{c}_{2}}\]. Put these equations in (3),
We get,
\[\Rightarrow \]\[\dfrac{{{y}^{1+1}}}{1+1}+{{c}_{1}}={{e}^{x}}+{{c}_{2}}\]
After simplification we get,
\[\dfrac{{{y}^{2}}}{2}={{e}^{x}}+{{c}_{{}}}\] where \[{{c}_{{}}}\]is the resultant constant of \[{{c}_{1}},{{c}_{2}}\]
 Now lets us multiply with 2 on both sides, we get
\[\Rightarrow \]\[2\times \dfrac{{{y}^{2}}}{2}=\left( {{e}^{x}}+{{c}_{{}}} \right)\times 2\]
After simplification we get
\[{{y}^{2}}=2{{e}^{x}}+c\]
Now apply root on both sides, we get
\[\Rightarrow \]\[\sqrt{{{y}^{2}}}=\sqrt{2{{e}^{x}}+c}\]………………(4)
As we know from the basic formula that \[\sqrt{{{a}^{2}}}=a\] ,we can write \[\sqrt{{{y}^{2}}}=y\]
So, equation (4) can be written as
\[\Rightarrow \]\[y=\sqrt{2{{e}^{x}}+c}\]…………………(5)
From the question we were also given that \[y\left( 0 \right)=1\]
\[y\left( 0 \right)=1\] this says that at \[x=0\] we get \[y=1\]
Now let us put \[x=0\] and \[y=1\]in equation (5)
We get,
\[\Rightarrow \]\[1=\sqrt{2{{e}^{0}}+c}\]
Now let us do squaring in both sides
\[\Rightarrow \]\[{{1}^{2}}={{\left( \sqrt{2{{e}^{0}}+c} \right)}^{2}}\]
After simplification we get
\[\Rightarrow \]\[1=2+c\] \[\left[ {{e}^{0}}=1 \right]\]
\[\Rightarrow \]\[c=-1\]
Now put \[c=-1\]in equation (5), we get
\[y=\sqrt{2{{e}^{x}}-1}\]
Therefore, \[y=\sqrt{2{{e}^{x}}-1}\] is the solution of given differential equation \[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\] with \[y\left( 0 \right)=1\].

Note: Students should do calculations correct to avoid errors in the final answer. Students should use proper formula .in case of misconception it will lead to a huge mistake while finding the solution of differential equation \[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\].