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Question

# In each of the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: (i) y = ex + 1 y'' − y' = 0 (ii) y = x2 + 2x + C y' − 2x − 2 = 0 (iii) y = cos x + C y' + sin x = 0 (iv) y = $\sqrt{1+{x}^{2}}$ y' = $\frac{xy}{1+{x}^{2}}$ (v) y = x sin x xy' = y + x $\sqrt{{x}^{2}-{y}^{2}}$ (vi) $y=\sqrt{{a}^{2}-{x}^{2}}$ $x+y\frac{dy}{dx}=0$

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Solution

## (i) We have, y'' − y' = 0 .....(1) Now, y = ex +1 $⇒y\text{'}={e}^{x}\phantom{\rule{0ex}{0ex}}⇒y\text{'}\text{'}={e}^{x}$ Putting the above values in (1), we get $\mathrm{LHS}={e}^{x}-{e}^{x}=0=\mathrm{RHS}$ Thus, y = ex +1 is the solution of the given differential equation. (ii) We have, y' − 2x − 2 = 0 .....(1) Now, y = x2 + 2x + C $⇒y\text{'}=2x+2$ Putting the above value in (1), we get $\mathrm{LHS}=2x+2-2x-2=0=\mathrm{RHS}$ Thus, y = x2 + 2x + C is the solution of the given differential equation. (iii) We have, y' + sin x = 0 .....(1) Now, y = cos x + C $⇒y\text{'}=-\mathrm{sin}x$ Putting the above value in (1), we get $\mathrm{LHS}=-\mathrm{sin}x+\mathrm{sin}x=0=\mathrm{RHS}$ Thus, y = cos x + C is the solution of the given differential equation. (iv) We have, y' = $\frac{xy}{1+{x}^{2}}$ .....(1) Now, y = $\sqrt{1+{x}^{2}}$ $⇒y\text{'}=\frac{x}{\sqrt{1+{x}^{2}}}$ Putting the above value in (1), we get $\mathrm{LHS}=\frac{x}{\sqrt{1+{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{x}{\sqrt{1+{x}^{2}}}×\frac{\sqrt{1+{x}^{2}}}{\sqrt{1+{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{xy}{1+{x}^{2}}=\mathrm{RHS}$ Thus, y = $\sqrt{1+{x}^{2}}$ is the solution of the given differential equation. (v) We have, xy' = y + x $\sqrt{{x}^{2}-{y}^{2}}$ .....(1) Now, y = x sin x $⇒y\text{'}=\mathrm{sin}x+x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}$ Putting the above value in (1), we get $\mathrm{LHS}=x\left(\mathrm{sin}x+x\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+{x}^{2}\mathrm{cos}x\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+x\left(x\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+x\left(x\sqrt{1-{\mathrm{sin}}^{2}x}\right)\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+x\left(\sqrt{{x}^{2}-{x}^{2}{\mathrm{sin}}^{2}x}\right)\phantom{\rule{0ex}{0ex}}=y+x\left(\sqrt{{x}^{2}-{y}^{2}}\right)=\mathrm{RHS}$ Thus, y = x sin x is the solution of the given differential equation. (v) We have, xy' = y + x $\sqrt{{x}^{2}-{y}^{2}}$ .....(1) Now, y = x sin x $⇒y\text{'}=\mathrm{sin}x+x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}$ Putting the above value in (1), we get $\mathrm{LHS}=x\left(\mathrm{sin}x+x\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+{x}^{2}\mathrm{cos}x\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+x\left(x\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+x\left(x\sqrt{1-{\mathrm{sin}}^{2}x}\right)\phantom{\rule{0ex}{0ex}}=x\mathrm{sin}x+x\left(\sqrt{{x}^{2}-{x}^{2}{\mathrm{sin}}^{2}x}\right)\phantom{\rule{0ex}{0ex}}=y+x\left(\sqrt{{x}^{2}-{y}^{2}}\right)=\mathrm{RHS}$ Thus, y = x sin x is the solution of the given differential equation. (vi) We have, $x+y\frac{dy}{dx}=0$ .....(1) Now, $y=\sqrt{{a}^{2}-{x}^{2}}$ $⇒y\text{'}=\frac{-x}{\sqrt{{a}^{2}-{x}^{2}}}\phantom{\rule{0ex}{0ex}}$ Putting the above value in (1), we get $\mathrm{LHS}=x+y\left(\frac{-x}{\sqrt{{a}^{2}-{x}^{2}}}\right)\phantom{\rule{0ex}{0ex}}=x+\sqrt{{a}^{2}-{x}^{2}}\left(\frac{-x}{\sqrt{{a}^{2}-{x}^{2}}}\right)\phantom{\rule{0ex}{0ex}}=x+\sqrt{{a}^{2}-{x}^{2}}\left(\frac{-x}{\sqrt{{a}^{2}-{x}^{2}}}\right)\phantom{\rule{0ex}{0ex}}=x-x=0=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$ Thus, $y=\sqrt{{a}^{2}-{x}^{2}}$ is the solution of the given differential equation.

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