# S9.3.a de with u subst

• MHB
• karush

#### karush

Gold Member
MHB
Solve the de $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$
sepate variables
$$\displaystyle \dfrac{dy}{\sqrt{y}\, \cos^2{\sqrt{y}} }=\dfrac{1}{7}\, dx \implies \int{\dfrac{{d}y}{\sqrt{y}\,\cos^2{\left(\sqrt{y} \right) }} } = \int{ \dfrac{1}{7}\,{d}x}$$
ok i think u subst is next ... maybe...
$$u=\sqrt{y} \therefore du=\dfrac{{d}y}{2\,\sqrt{y}}$$

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Solve the de $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$
sepate variables
$$\displaystyle \dfrac{dy}{\sqrt{y}\, \cos^2{\sqrt{y}} }=\dfrac{1}{7}\, dx \implies \int{\dfrac{{d}y}{\sqrt{y}\,\cos^2{\left(\sqrt{y} \right) }} } = \int{ \dfrac{1}{7}\,{d}x}$$
ok i think u subst is next ... maybe...
$$u=\sqrt{y} \therefore du=\dfrac{{d}y}{2\,\sqrt{y}}$$
You stopped too soon! What did you get when you did the substitution?

-Dan

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using your sub $$u = \sqrt{y}$$ ...

$$\displaystyle 2 \int \dfrac{1}{\cos^2{\sqrt{y}}} \cdot \dfrac{dy}{2\sqrt{y}} = 2\int \sec^2{u} \, du$$