# Exact differential equation problem 5 [closed]

$$(te^{tx}+2x)\frac{dx}{dt}+xe^{xt}=0$$ I found that this is an exact equation by parial derevatives,please can anybody tell me the way to solve this..

## closed as off-topic by Dmoreno, user91500, Namaste, Frits Veerman, kingW3Jun 22 '17 at 13:10

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• If you found out that you have an exact equation, what prevents you from solving it the way your book (which I assume you have got) explains? – mickep Jun 21 '17 at 5:50
• What you tried? Do you have any of your own thoughts? If this is the fifth problem relating to exact differential equations then how have the other four been solved? – Rumplestillskin Jun 21 '17 at 5:56

$$\psi(x,t)$$

$$\frac{\mathbb{d}}{\mathbb{d}t}\psi(x(t),t)=0$$

$$\frac{\partial \psi }{\partial x}\frac{\mathbb{d}x}{\mathbb{d}t}+\frac{\partial \psi }{\partial t}=0$$

$$\frac{\partial \psi }{\partial t}=xe^{xt}$$

$$\psi=e^{xt}+f(x)$$

$$\frac{\partial \psi}{\partial x}=te^{xt}+f'(x)=te^{xt}+2x$$

$$f'(x)=2x$$

$$f(x)=x^2$$

$$\psi = k =e^{xt}+x^2$$

$$k=e^{x(t)t}+x(t)^2$$

suppose we are given an initial condition x(0) $$k=1+x(0)^2$$

$$e^{x(t)t}+x(t)^2=1+x(0)^2$$

$$(te^{tx}+2x)dx+xe^{xt}dt=0$$ @ channa jayasanka : You already found that it is an exact differential. So, you just have to apply the method taught at school to solve the ODE. May be, you don't know how to do it :

Let $F(x,t)$ be the function which the total derivative is $$dF=(te^{tx}+2x)dx+xe^{xt}dt \quad\to\quad \begin{cases} \frac{\partial F}{\partial x}=te^{tx}+2x \\ \frac{\partial F}{\partial t}=xe^{xt} \end{cases}$$ Integration with respect to $x$ only : $\quad F=\int (te^{tx}+2x)dx=e^{tx}+x^2+f(t)$

Integration with respect to $t$ only : $\quad F=\int xe^{xt}dt=e^{tx}+g(x)$ $$F=e^{tx}+x^2+f(t)=e^{tx}+g(x) \quad\to\quad \begin{cases}f(t)=0 \\ g(x)=x^2\end{cases} \quad\to\quad F=e^{tx}+x^2$$ Since $dF=0 \quad\to\quad F=C=$ constant $\quad\to\quad$ $\boxed{e^{tx}+x^2=C}$

$\\$

An alternative way is the change of function : $\quad e^{xt}=y(x) \quad\to\quad t=\frac{\ln(y)}{x}\quad\to\quad \frac{dt}{dx}=\frac{1}{xy}\frac{dy}{dx}-\frac{\ln(y)}{x^2} =\frac{1}{xy}\frac{dy}{dx}-\frac{t}{x}$ $$(te^{xt}+2x)+xe^{xt}\left(\frac{1}{xy}\frac{dy}{dx}-\frac{t}{x}\right)=(ty+2x)+xy\left(\frac{1}{xy}\frac{dy}{dx}-\frac{t}{x}\right)=0$$ After simplification : $\qquad 2x+\frac{dy}{dx}=0 \quad\to\quad x^2+y=C$

The solution expressed on the form of implicit equation is : $$x^2+e^{xt}=C$$ Or, on explicit form: $$t=\frac{\ln(C-x^2)}{x}$$