$$ MATH $131$

HW$1$

Solve the differential equation

$(tanx)\frac{dy}{dx}+(ta{n}^{2}x)y=2$

Solve the following equation with the initial value $y(0)=3.$

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

Solve equation $\frac{dy}{dx}=\frac{x+e^x}{yta{n}^{-1}y}.$ Reminder: $ta{n}^{-1}$ denotes arctan$,$ not the reciprocal of the tangent function.

Solve equation $\frac{dy}{dx}=\frac{4x^2+3y^2}{2xy}.$

Hard integral: Solve ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{(1-x^3{)}^{\frac{1}{3}}}$

If you haven't seen a similar integral before, it is very unclear what to do$.$ But, believe it or not, the following substitution magically transforms the integral into something doable. There may be other ways to do this, but a method I know that works is the substitution

$x=\frac{u}{(1+u^3{)}^{\frac{1}{3}}}$

When you work out the integral involving $u,$ at the last step you may want to consult a table or even technology to check your answer, but please show me a fully written solution.

$6.$ Much easier integral $($easier if you use the right method$)$: Solve ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{x-8x^4}.$ Hint: Use the trick I showed you in class. When you have an integral of the form ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{A+B{x}^a}$ for numbers $A,B,a,$ the substitution $u=y^{1-a}$ leads to a very easy solution. Here, of course, $a$ is $4.$

$1$