# Solving a Non linear ODE

Can the following differential equation be solved analytically by some method

$$v(v’’’ + 3v’’ +52v’ + 50v) + 4(v’’’ + 3 v’’ + 52v’ + 100v) = 0,\\ v(0) = v’(0) = v’’(0) = 1.$$

The solutions to each linear DE in the parenthesis is known. On observation one can see that for the first DE the roots of characterstic equation are $$-1, -1\pm7i$$. And the roots to the other are $$-2,-0.5\pm i\sqrt{49.75}$$. I have tried solving it numerically on matlab and the solution looks like a sinusoidal wave squeezed between 2 exponentially decaying functions. And the solution decays to zero. Presently, I am trying if functions of the following form might be able to fit into the solution $$v(x) = \alpha - \sqrt{\alpha^2 + (a_1 e^{-x}+ a_2e^{-2x})+(b_1e^{-x}+b_2e^{-2x})(A\cos\delta x + B\sin\delta x)}$$

• Can rewrite as $(v+4)(v'''+3v''+52v'+50v)+200v=0.$ Don't know if that gains you anything. – Adrian Keister Jun 7 at 19:43

Hint:

$$v(v'''+3v''+52v'+50v)+4(v'''+3v''+52v'+100v)=0$$ with $$v(0)=v'(0)=v''(0)=1$$

$$(v+4)(v'''+3v''+52v'+50v)+200v=0$$ with $$v(0)=v'(0)=v''(0)=1$$

Let $$u=v+4$$ ,

Then $$u(u'''+3u''+52u'+50(u-4))+200(u-4)=0$$ with $$u(0)=5$$ , $$u'(0)=u''(0)=1$$

$$u(u'''+3u''+52u'+50u)-800=0$$ with $$u(0)=5$$ , $$u'(0)=u''(0)=1$$

$$u'''+3u''+52u'+50u-\dfrac{800}{u}=0$$ with $$u(0)=5$$ , $$u'(0)=u''(0)=1$$

Let $$u=e^{-x}w$$ ,

Then $$u'=e^{-x}w'-e^{-x}w$$

$$u''=e^{-x}w''-e^{-x}w'-e^{-x}w'+e^{-x}w=e^{-x}w''-2e^{-x}w'+e^{-x}w$$

$$u'''=e^{-x}w'''-e^{-x}w''-2e^{-x}w''+2e^{-x}w'+e^{-x}w'-e^{-x}w=e^{-x}w'''-3e^{-x}w''+3e^{-x}w'-e^{-x}w$$

$$\therefore e^{-x}w'''-3e^{-x}w''+3e^{-x}w'-e^{-x}w+3e^{-x}w''-6e^{-x}w'+3e^{-x}w+52e^{-x}w'-52e^{-x}w+50e^{-x}w-\dfrac{800}{e^{-x}w}=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$e^{-x}w'''+49e^{-x}w'-\dfrac{800}{e^{-x}w}=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$ww'''+49ww'-800e^{2x}=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$\int ww'''~dx+49\int ww'~dx-800\int e^{2x}~dx=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$\int w~dw''+49\int w~dw-800\int e^{2x}~dx=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$ww''-\int w''~dw+49\int w~dw-800\int e^{2x}~dx=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$ww''-\int w'w''~dx+49\int w~dw-800\int e^{2x}~dx=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$ww''-\int w'~dw'+49\int w~dw-800\int e^{2x}~dx=0$$ with $$w(0)=5$$ , $$w'(0)=6$$ , $$w''(0)=18$$

$$ww''-\dfrac{(w')^2}{2}+\dfrac{49w^2}{2}-400e^{2x}=933$$ with $$w(0)=5$$ , $$w'(0)=6$$

Let $$w=z^2$$ ,

Then $$w'=2zz'$$

$$w''=2zz''+2(z')^2$$

$$\therefore2z^3z''+2z^2(z')^2-2z^2(z')^2+\dfrac{49z^4}{2}-400e^{2x}=933$$ with $$z(0)=\pm\sqrt5$$ , $$z'(0)=\pm\dfrac{3}{\sqrt5}$$

$$2z^3z''+\dfrac{49z^4}{2}=400e^{2x}+933$$ with $$z(0)=\pm\sqrt5$$ , $$z'(0)=\pm\dfrac{3}{\sqrt5}$$

$$z''+\dfrac{49z}{4}=\dfrac{400e^{2x}+933}{2z^3}$$ with $$z(0)=\pm\sqrt5$$ , $$z'(0)=\pm\dfrac{3}{\sqrt5}$$

• But still the resulting equation is non-linear and cannot be solved by usual techniques, right? – breaku Jun 8 at 17:13