Question

solve: e^dy/dx = x^2​

Answer 1

Concept

The technique of finding the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative is known as integration by parts or partial integration.

Find

$e^{^{dy}/dx} = x^2$

Solution

Taking ln on both sides

$ln\ e^{dy/dx} = ln\ x^2\\dy/dx\ . ln\ e= 2, ln\ x\\dy/dx\ = 2, ln\ x\\dy = 2. ln\ x.\ dx \ \ \ \ .......(1)$

Now, we know that;

$\int\limits {(u.v)} \, dx = u \int\limits {v} \, dx\ - \int\limits [{\frac{du}{dx} \int\limits {v} \, dx]} \, dx$

solving (1).

$dy = 2. ln\ x.\ dx \\\\Integrating\ on\ both\ sides;\\\int\limits\ dy = 2. \int\limits\ ln\ x.\ dx \\y = 2.[ ln\ x . \int\limits\ dx\ - \int\limits\ (\frac{1}{x} \int\limits\ dx )\ dx \\y = 2. [ln\ x . (x)\ - \int\limits\ \frac{1}{x} (x)\ dx]\\y= 2.[ x.\ ln\ x - x]\ + C \\y = 2x [ ln\x -1]\ + C$

Answer 2

Answer:

The solution of the given differential equation is

$y = x( log_{e}(x) - 1)+c$

Step-by-step explanation:

Given equation is

$e {}^{ \frac{dy}{dx} } = {x}^{2}$

Taking log on both sides we get

$log_{e}(e {}^{ \frac{dy}{dx} } ) = log_{e}(x {}^{2} ) \\ = > \frac{dy}{dx} = 2 log_{e}(x) \\ = > dy = 2 log_{e}(x) dx$

Integrating on both sides we get

$\int \: dy = \int2 log_{e}(x) dx \\ = > y = 2 \int log_{e}(x) dx$

We solve the above integral by using ILATE Method which is

$\int \: udv = uv - \int \: vdu$

Similarly for

$\int \: log_{e}(x) dx$

we have

u=logx and dv=dx

$\int log_{e}(x) dx = log_{e}(x) x - \int \: x. \frac{1}{x} dx \\ = x log_{e}(x) - x = x( log_{e}(x) - 1)$

Therefore solution of our equation becomes

$y = 2x( log_{e}(x) - 1)+c$

#SPJ3