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Question

Question: The solution of \(\dfrac{{dy}}{{dx}} + y = {e^x}\) \( a)\,2y = {e^{2x}} + c \\\ b)\,2y{e^x...

The solution of dydx+y=ex\dfrac{{dy}}{{dx}} + y = {e^x}
a)2y=e2x+c b)2yex=ex+c c)2yex=e2x+c d)2ye2x=2ex+c  a)\,2y = {e^{2x}} + c \\\ b)\,2y{e^x} = {e^x} + c \\\ c)\,2y{e^x} = {e^{2x}} + c \\\ d)\,2y{e^{2x}} = 2{e^x} + c \\\

Explanation

Solution

In this type of question you have to find out the (IF) that is integrating factor and then use integration to solve this problem.

Complete step-by-step answer:
So question is dydx+y=ex\dfrac{{dy}}{{dx}} + y = {e^x}
Here dydx\dfrac{{dy}}{{dx}}is the derivative of yy with respect to xx
So we know if the equation is dydx+f(x)y=g(x)\dfrac{{dy}}{{dx}} + f(x)y = g(x)
Then Integrating Factor (IF) =ef(x)dx = {e^{\int {f(x)dx} }}
And then the general solution is given by
(IF)y=(IF)g(x)dx(IF)y = \int {(IF)g(x)dx}
Now, let us find the (IF) of given equation dydx+y=ex\dfrac{{dy}}{{dx}} + y = {e^x}
Here f(x)=1&g(x)=exf(x) = 1\,\,\,\,\,\& \,\,\,\,g(x) = {e^x}
(IF) =ef(x)dx = {e^{\int {f(x)dx} }}
=edx= {e^{\int {dx} }}
And we know dx=x\int {dx = x}
So we get (IF) =ex = {e^x}
Now we know the general solution is given by
(IF)y=(IF)g(x)dx(IF)y = \int {(IF)g(x)dx}
Here,
exy=exexdx exy=e2xdx  {e^x}y = \int {{e^x}{e^x}dx} \\\ {e^x}y = \int {{e^{2x}}dx} \\\
And we know eaxdx=eaxa+c\int {{e^{ax}}dx = \dfrac{{{e^{ax}}}}{a}} + c , using this we get
exy=e2x2+c,{e^x}y = \dfrac{{{e^{2x}}}}{2} + {c^,}
2exy=e2x+2c, 2exy=e2x+c,,  2{e^x}y = {e^{2x}} + 2{c^,} \\\ 2{e^x}y = {e^{2x}} + {c^{,,}} \\\

So, the correct answer is “Option C”.

Note: This type of differential equation is solved by finding the integration factor and putting into the formula for equation of differential equation, if differential equation is given by dydx+f(x)y=g(x)\dfrac{{dy}}{{dx}} + f(x)y = g(x) then (IF) =ef(x)dx = {e^{\int {f(x)dx} }}and general solution is given by (IF)y=(IF)g(x)dx(IF)y = \int {(IF)g(x)dx}