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Test:	DELA - 2017 Spring - Midterm I
Number of Questions:	9
Time:

$\mbox{Solve} \ y^{\prime}-\dfrac{4}{x}y=x^5\sin x, \ x>0.$

	$y=x^4(-x \cos x+\sin x)+c$
	$\mbox{None of the above}$
	$y=x^4(-x \cos x+\sin x+c)$
	$y=x^4(-x \cos x-\sin x+c)$
	$y=x^4(-x \cos x-\sin x)+c$

$\mbox{Solve} \ e^{x+y}y^{\prime}=1.$

	$e^y+e^{-x}=c$
	$e^y+e^{x}=c$
	$e^y-e^{-x}=c$
	$\mbox{None of the above}$
	$-e^y+e^{x}=c$
	$-e^y+e^{-x}=c$

$\mbox{Solve} \ (4e^{2x}+2xy-y^2)dx+(x-y)^2dy=0.$

	$6e^{2x}-3x^2y-3xy^2+y^3=c$
	$\mbox{None of the above}$
	$6e^{2x}+3x^2y+3xy^2+y^3=c$
	$6e^{2x}+3x^2y-3xy^2-y^3=c$
	$6e^{2x}+3x^2y-3xy^2+y^3=c$
	$6e^{2x}-3x^2y-3xy^2-y^3=c$

$\mbox{Solve} \ x^2ydx+y(x^3+e^{-3y}\sin y)dy=0.$

	$\mbox{None of the above}$
	$\dfrac{x^3e^{-3y}}{3}+\cos y=c$
	$\dfrac{x^3e^{3y}}{3}-\cos y=c$
	$\dfrac{x^3e^{3y}}{3}+\cos y=c$
	$\dfrac{x^3e^{-3y}}{3}-\cos y=c$

$\hspace{-6mm} \mbox{Compute the third Picard iteration for the initial-value problem}\\ \dfrac{dx}{dt}=2x, \ x(0)=3.$

	$x_3(t)=3+2t(2t^2+3t+3)$
	$\mbox{None of the above}$
	$x_3(t)=3+2t(2t^2+3t+2)$
	$x_3(t)=3+3t(2t^2+3t+3)$
	$x_3(t)=3+2t(3t^2+2t+3)$

$\hspace{-6mm} \mbox{Solve the initial-value problem}\\ y^{\prime\prime}-4y^{\prime}+5y=0, \ y(0)=3, \ y^{\prime}(0)=5.$

	$y(x)=e^{2x}(3\sin x+\cos x)$
	$y(x)=e^{2x}(3\cos x-\sin x)$
	$y(x)=e^{2x}(3\sin x-\cos x)$
	$\mbox{None of the above}$
	$y(x)=e^{2x}(3\cos x+\sin x)$

$\mbox{Solve} \ y^{\prime\prime}+6y^{\prime}+9y=\dfrac{2e^{-3x}}{x^2+1}.$

	$\mbox{None of the above}$
	$y(x)=e^{-3x}[c_1+c_2x+2x\tan^{-1}x-\ln(x^2-1)]$
	$y(x)=e^{-3x}[c_1+c_2x-2x\tan^{-1}x-\ln(x^2-1)]$
	$y(x)=e^{-3x}[c_1+c_2x+2x\tan^{-1}x+\ln(x^2+1)]$
	$y(x)=e^{-3x}[c_1+c_2x+2x\tan^{-1}x-\ln(x^2+1)]$

$\mbox{Solve} \ y^{\prime\prime}+16y=4\cos x.$

	$\mbox{None of the above}$
	$y(x)=c_{1}\cos 4x+c_{2}\sin 4x+\dfrac{16}{15}\cos x$
	$y(x)=c_{1}\cos 4x+c_{2}\sin 4x+\dfrac{4}{15}\cos x$
	$y(x)=c_{1}\cos 4x+c_{2}\sin 4x+\dfrac{8}{15}\cos x$
	$y(x)=c_{1}\cos 4x+c_{2}\sin 4x+\dfrac{2}{15}\cos x$

$\hspace{-6mm} \mbox{Solve the initial-value problem}\\ y^{\prime\prime}+t^2y^{\prime}+2ty=0; \ y(0)=1, \ y^{\prime}(0)=0 .$

	$y(t)=\sum_{n=0}^{\infty}\dfrac{t^{3n}}{3^nn!}$
	$y(t)=\sum_{n=0}^{\infty}\dfrac{(-1)^nt^{2n}}{n!}$
	$\mbox{None of the above}$
	$y(t)=\sum_{n=0}^{\infty}\dfrac{t^{2n}}{2^nn!}$
	$y(t)=\sum_{n=0}^{\infty}\dfrac{(-1)^nt^{3n}}{3^n}$
	$y(t)=\sum_{n=0}^{\infty}\dfrac{(-1)^nt^{3n}}{3^nn!}$
	$y(t)=\sum_{n=0}^{\infty}\dfrac{(-1)^nt^{2n}}{2^nn!}$

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