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# Verify Solutions of DEs Exercises Worked 2

Exercises

## Exercises Continued

Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function

### 4.

{$\small y=\left(1+ce^{-x^2/2}\;\right) \left(1-ce^{-x^2/2}\;\right)^{-1}\;;\; 2y'+x(y^2-1)=0$}

Some preliminaries to make things easier:

{\begin{align} \left(1+ce^{-x^2/2}\;\right)^\prime &= 0+c({{-x^2}\over{2}})^\prime (e^{-x^2/2}\;)^\prime \cr &= c\left(-{1\over 2}\right)(2x)(e^{-x^2/2}\;) \cr &= -cxe^{-x^2/2} \cr \left(1-ce^{-x^2/2}\;\right)^\prime &= cxe^{-x^2/2}\end{align}}

{\begin{align} y^2&= \left[\left(1+ce^{-x^2/2}\;\right) \left(1-ce^{-x^2/2}\;\right)^{-1}\right]^2 \cr &= \left(1+ce^{-x^2/2}\;\right)^2 \left(1-ce^{-x^2/2}\;\right)^{-2} \cr &= {{1+ 2ce^{-x^2/2} + ce^{-x^2} }\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} \end{align}}

{\begin{align} y^\prime &= {(-cxe^{-x^2/2}\;) \left(1-ce^{-x^2/2}\;\right) -(cxe^{-x^2/2}\;)\left(1+ce^{-x^2/2}\;\right)\over{ \left(1-ce^{-x^2/2}\;\right)^2 }} \cr &= {(-cxe^{-x^2/2}\;)- (-cxe^{-x^2/2}\;)(ce^{-x^2/2}\;) -\left((cxe^{-x^2/2}\;)+(cxe^{-x^2/2}\;)(ce^{-x^2/2}\;)\right)\over{ \left(1-ce^{-x^2/2}\;\right)^2 }} \cr &= { {-cxe^{-x^2/2} + cxe^{-x^2} - cxe^{-x^2/2} - cxe^{-x^2} }\over{ \left(1-ce^{-x^2/2}\;\right)^2 }} \cr &= { {-2cxe^{-x^2/2} }\over{ \left(1-ce^{-x^2/2}\;\right)^2 }} \end{align}}

Now for the main course:

{\begin{align} 2y'+x(y^2-1)&=0 \cr 2 { {-2cxe^{-x^2/2} }\over{ \left(1-ce^{-x^2/2}\;\right)^2 }} + x {{1+ 2ce^{-x^2/2} + ce^{-x^2} }\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} -x \quad & ? \quad 0 \cr { {-4cxe^{-x^2/2} }\over{ \left(1-ce^{-x^2/2}\;\right)^2 }} + {{x+ 2cxe^{-x^2/2} + cxe^{-x^2} }\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} -x \quad & ? \quad 0 \cr { {x- 2cxe^{-x^2/2} + cxe^{-x^2} }\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} - {{x(1-ce^{-x^2/2})^{2}}\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} \quad & ? \quad 0 \cr { {x- 2cxe^{-x^2/2} + cxe^{-x^2} }\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} - {{(x-2cxe^{-x^2/2}-xce^{-x^2})}\over{ \left(1-ce^{-x^2/2}\;\right)^{2}}} \quad & ? \quad 0 \cr \therefore 0 &= 0\end{align}}

### 5.

{$$\large y={\tan\left( {x^3\over3}+c\right)}; y'=x^2(1+y^2)$$}

{\begin{align} y^\prime &= \left[\tan\left( {x^3\over3}+c\right)\right]^\prime \cr y^\prime &= \left({1\over 3 }x^3+c\right)^\prime \left(\tan^\prime\left( {x^3\over3}+c\right)\right) \cr y^\prime &= x^2 \sec^2\left( {x^3\over3}+c\right) \cr y^\prime &= x^2 (1 + \tan^2\left( {x^3\over3}+c\right) ) \cr \therefore y^\prime &= x^2 (1 + y^2 ) \end{align}}

### 6.

{$$\large y=(c_1+c_2x)e^x+\sin x+x^2\;; y''-2y'+y=-2 \cos x+x^2-4x+2$$}

{\begin{align} y&=(c_1+c_2x)e^x+\sin x+x^2 \cr y&=c_1e^x+c_2xe^x+\sin x+x^2 \cr y^\prime &=c_1e^x +c_2e^x +c_2xe^x+\cos x+ 2x \cr 2y^\prime &=2c_1e^x +2c_2e^x +2c_2xe^x+2\cos x+ 4x \cr y^{\prime\prime} &=c_1e^x +2c_2e^x +c_2xe^x-\sin x+ 2 \cr \end{align}}

 {$y^{\prime\prime}$} {$+c_1e^x$} {$+2c_2e^x$} {$+c_2xe^x$} {$-\sin x$} {} {} {} {$+2$} {$-2y^\prime$} {$-2c_1e^x$} {$-2c_2e^x$} {$-2c_2xe^x$} {$-2\cos x$} {} {$-4x$} {$y$} {$c_1e^x$} {$+c_2xe^x$} {$+\sin x$} {$+ x^2$} {$y^{\prime\prime}-2y^\prime +y$} = {$-2\cos x$} {$+ x^2$} {$-4x$} {$+2$}

Sources:

Trench, William F., "Elementary Differential Equations with Boundary Value Problems" (2013). Faculty Authored Books. 9. Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

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