Step 1

\(\displaystyle\text{We have: }\ {f}{''}{\left({x}\right)}={\sin{{x}}},{f}'{\left({0}\right)}={1},{f{{\left({0}\right)}}}={6}\)

\(\displaystyle\text{We integrate }\ {f}{''}{\left({x}\right)}={\sin{{x}}}\ \text{ (first) and }\ {f}'{\left({0}\right)}\ \text{ (second) to obtain a generalized solution:}\)

\(\displaystyle\int{f}{''}{\left({x}\right)}{\left.{d}{x}\right.}=\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}+{C}_{{1}}\)

Step 2

Now, we are applying the initial condition \(\displaystyle{f}'{\left({0}\right)}={1}:\)

\(\displaystyle-{\cos{{0}}}+{C}_{{1}}={1}\)

\(\displaystyle-{1}+{C}_{{1}}={1}\)

\(\displaystyle{C}_{{1}}={1}+{1}\)

\(\displaystyle{C}_{{1}}={2}\)

\(\displaystyle\text{So: }\ {f}'{\left({x}\right)}=-{\cos{{x}}}+{2}\)

Now, we integrate f`(x):

\(\displaystyle\int{f}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{\left(-{\cos{{x}}}+{2}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle=\int-{\cos{{x}}}{\left.{d}{x}\right.}+\int{2}{\left.{d}{x}\right.}\)

\(\displaystyle=-{\sin{{x}}}+{2}{x}+{C}_{{2}}\)

Now, we are applying the initial condition f(0)=6:

\(\displaystyle{\sin{{0}}}+{2}\cdot{0}+{C}_{{2}}={6}\)

\(\displaystyle{0}+{C}_{{2}}={6}\)

\(\displaystyle{C}_{{2}}={6}\)

\(\displaystyle\text{So: }\ {f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)

Finaly, the particular solution is:

\(\displaystyle{f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)

Step 3

Explanetion:

\(\displaystyle{\left({1}\right)}:{f}{''}{\left({x}\right)}={\sin{{x}}}\)

\(\displaystyle{\left({2}\right)}:\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}+{C}\)

\(\displaystyle{\left({3}\right)}:{f}'{\left({x}\right)}=-{\cos{{x}}}+{2}\)

\(\displaystyle{\left({4}\right)}:\int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}+{C}\)

Result

\(\displaystyle{f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)

\(\displaystyle\text{We have: }\ {f}{''}{\left({x}\right)}={\sin{{x}}},{f}'{\left({0}\right)}={1},{f{{\left({0}\right)}}}={6}\)

\(\displaystyle\text{We integrate }\ {f}{''}{\left({x}\right)}={\sin{{x}}}\ \text{ (first) and }\ {f}'{\left({0}\right)}\ \text{ (second) to obtain a generalized solution:}\)

\(\displaystyle\int{f}{''}{\left({x}\right)}{\left.{d}{x}\right.}=\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}+{C}_{{1}}\)

Step 2

Now, we are applying the initial condition \(\displaystyle{f}'{\left({0}\right)}={1}:\)

\(\displaystyle-{\cos{{0}}}+{C}_{{1}}={1}\)

\(\displaystyle-{1}+{C}_{{1}}={1}\)

\(\displaystyle{C}_{{1}}={1}+{1}\)

\(\displaystyle{C}_{{1}}={2}\)

\(\displaystyle\text{So: }\ {f}'{\left({x}\right)}=-{\cos{{x}}}+{2}\)

Now, we integrate f`(x):

\(\displaystyle\int{f}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{\left(-{\cos{{x}}}+{2}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle=\int-{\cos{{x}}}{\left.{d}{x}\right.}+\int{2}{\left.{d}{x}\right.}\)

\(\displaystyle=-{\sin{{x}}}+{2}{x}+{C}_{{2}}\)

Now, we are applying the initial condition f(0)=6:

\(\displaystyle{\sin{{0}}}+{2}\cdot{0}+{C}_{{2}}={6}\)

\(\displaystyle{0}+{C}_{{2}}={6}\)

\(\displaystyle{C}_{{2}}={6}\)

\(\displaystyle\text{So: }\ {f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)

Finaly, the particular solution is:

\(\displaystyle{f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)

Step 3

Explanetion:

\(\displaystyle{\left({1}\right)}:{f}{''}{\left({x}\right)}={\sin{{x}}}\)

\(\displaystyle{\left({2}\right)}:\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}+{C}\)

\(\displaystyle{\left({3}\right)}:{f}'{\left({x}\right)}=-{\cos{{x}}}+{2}\)

\(\displaystyle{\left({4}\right)}:\int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}+{C}\)

Result

\(\displaystyle{f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)