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Question: What is a solution to the differential equation \[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}\] with \[y\lef...

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

Explanation

Solution

For the equation dydx=exy\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 ∫exdx=ex+c\int{{{e}^{x}}dx}={{e}^{x}}+c and ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c(where cc is constant) Before solving this question.

Complete step by step solution:
From the given question, we were given that
\Rightarrow $$$$\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y} with y(0)=1y\left( 0 \right)=1
Consider equation dydx=exy\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}……………….(1)
Now we have to cross multiply the equation, we get
\Rightarrow $$$$y\times dy={{e}^{x}}\times dx……………(2)
As we separated all the yyterms to one side and xx to other side, we can use variable separable methods now.
Now apply integration on both sides. We get
\Rightarrow $$$$\int{y}\times dy=\int{{{e}^{x}}}\times dx……………..(3)
As we know from the basic integration formulas ∫exdx=ex+c\int{{{e}^{x}}dx}={{e}^{x}}+c and ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c(where cc is constant) we can write ∫y1dx=y1+11+1+c1\int{{{y}^{1}}dx=\dfrac{{{y}^{1+1}}}{1+1}}+{{c}_{1}} and ∫ex×dx=ex+c2\int{{{e}^{x}}}\times dx={{e}^{x}}+{{c}_{2}}. Put these equations in (3),
We get,
\Rightarrow $$$$\dfrac{{{y}^{1+1}}}{1+1}+{{c}_{1}}={{e}^{x}}+{{c}_{2}}
After simplification we get,
y22=ex+c\dfrac{{{y}^{2}}}{2}={{e}^{x}}+{{c}_{{}}} where c{{c}_{{}}}is the resultant constant of c1,c2{{c}_{1}},{{c}_{2}}
Now lets us multiply with 2 on both sides, we get
\Rightarrow $$$$2\times \dfrac{{{y}^{2}}}{2}=\left( {{e}^{x}}+{{c}_{{}}} \right)\times 2
After simplification we get
y2=2ex+c{{y}^{2}}=2{{e}^{x}}+c
Now apply root on both sides, we get
\Rightarrow $$$$\sqrt{{{y}^{2}}}=\sqrt{2{{e}^{x}}+c}………………(4)
As we know from the basic formula that a2=a\sqrt{{{a}^{2}}}=a ,we can write y2=y\sqrt{{{y}^{2}}}=y
So, equation (4) can be written as
\Rightarrow $$$$y=\sqrt{2{{e}^{x}}+c}…………………(5)
From the question we were also given that y(0)=1y\left( 0 \right)=1
y(0)=1y\left( 0 \right)=1 this says that at x=0x=0 we get y=1y=1
Now let us put x=0x=0 and y=1y=1in equation (5)
We get,
\Rightarrow $$$$1=\sqrt{2{{e}^{0}}+c}
Now let us do squaring in both sides
\Rightarrow $$$${{1}^{2}}={{\left( \sqrt{2{{e}^{0}}+c} \right)}^{2}}
After simplification we get
\Rightarrow $$$$1=2+c [e0=1]\left[ {{e}^{0}}=1 \right]
\Rightarrow $$$$c=-1
Now put c=−1c=-1in equation (5), we get
y=2ex−1y=\sqrt{2{{e}^{x}}-1}
Therefore, y=2ex−1y=\sqrt{2{{e}^{x}}-1} is the solution of given differential equation dydx=exy\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y} with y(0)=1y\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 dydx=exy\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{y}.